According to the Latest syllabus of Kashmir University A TEXT BOOK OF ELEMENTARY CALCULUS For 1st year of Three-year Degree Course By Prof. G.M. SHAH M.Sc., L.L.B. (Aligarh) Department of Mathematics A.S. College, Srinagar ALI MOHAMMAD & SONS SRINAGAR Published by Ali Mohammad & Sons Srinagar — Kasbrnh © Publishers Read These Books also : — 1. A Text Book of Higher Algebra For T.D.C. Part / ■2. A Text Book of Co-ordinate Geometry For T.D.C. Part / Price : Rs. Six only Undertaken through : V.V. Enterprisers Composed by : Deepak Printing Service Printed at : Mehta Printers, Delhi-l 10006 PREFACE TO THE THIRD EDITION The present edition has enabled me lo completely overhaul i^rhnn?ir%°V adding tO It Chapter IX on the Statement of Maclaurin’s Theorem and its applications to Simple Expansions.” Up-to-date University papJJs have also been added to enable the student to know what Lr^ of questions are usually set at the university level. It is, therefore, hoped that the book in its present form will prove more uselul to the students than before. Srinagar February, 1978 G.M. Shah PREFACE TO THE SECOND EDITION The book has been thoroughly revised and Theory of Sets has been included in the first chapter in accordance with the new syllabus of the Kashmir University. Integration as the limit of a has also been added to tlie chapter on Intcgriiiion. Srinagar July. 1968 G.M. Shah. PREFACE TO THE FIRST EDITION The cliief aim in writing this book has been to provide a suitable text-book on Calculus for the use of students studying in the hrst year of three-year degree course. It has been written in strict accordance with the syllabus prescribed by the J. K University. Every attempt has been made to make the concepts important topics such as “Limits” Differentiation , etc., have been dealt with in detail, and special care has been taken to deal with the difficulties of students of average ability. Various articles have been explained and illustrated by means of a number of solved examples. In short, no pains have been spared to present the subject matter in as lucid a manner as possible. inauMci received and' aSL^wleSgeT’’™'''™'"* “'""kfally Srinagar August, 1965 G.M. Shah SYLLABUS OF THE UNIVERSITY OF KASHMIR For 1st Year T.D.C, MATHEMATICS PAPER— II For 1978 onwards CALCULUS Section B (1) Elemcniary Functions such as : — Polynomial Functions, Rational functions. Trigonometric and inverse trigono- metric functions. Exponential and logarithmic functions. Hyperbolic functions and Inverse hyperbolic functions. Section C (1) Intuitive notion of the limit of a function. (2) Derivative Algebra and derivable functions. (3) Differentiation of algebraic, trigonometric, inverse trigonometric, exponential, logarithmic and hyperbolic functions. (4) The derivative as rale of change. (5) Dervative as tangent-slope. Applications to tangents and normals in Cartesian co-ordinates. Section D (1) Successive differentiation. Leibnitz rule. (2) Statement of Maclaurin’s Theorem with applications to simple expansions. (3) Definition of integral as limit of a sum with very simple applications. '=4 Ar =0 >-=1 .v=3 y— 16 etc. Evidently, the values of y depend upon the values of x. There- fore, we can say that y is a/uwe/Zon of A', while X is known as the ixrgumcnt. This relation is symbolically expressed as Hi) Consider y— tan x If .v=^ ,-=1 y= ^3 and so on. Here also we find that the value of y or tan x depends upon the value of x. Hence y=tan x is a function of x. Note. (/) Different symbols such as /tx), F(x), ^(x), etc., are used to represent functions when more than one function are ainder consideration. Hi) A function of more than one variable is denoted as/tx, y), .v), /(X, y, z), etc. 12. I f/{x) be a function of x, we can find its value for a given value of the argument .v. The value of/(x) for x—a is denot- . 2 d by/(a). -Ond Thus, if /(.v) = .va+l, f{ a) = oT- -I- 1 /(1)=2 /(0)-I we have for x=fl forx=l for.v=0 flidicn X is replaced by .V CONCEPTION OF A FUNCTION 3 1*3. Geometrical Representation of a Function The relation between a function and its argument is geometric cally represented by a curve. This curve is called the graph of the said function. >»=/(x) be a function of.x. We can get different values of >» corresponding to different values of at. Let us take the values of X as the abscissae and those ofyasthe ordinates and plot the points. Then a curve through these points is known as the graph of y=f{x). Example. Draw the graph of >'=3.y+5. Sol. The table of values of x and y is given below : 0 -1 1 1 1 -2 -4 1 1 -3 5 3 8 -1 -7 -4 The graph is a straight line which has been obtained by plotting the points (0. 5). (-1,2) (I, 8), (-2. -1) (-4. -7) and (-3, -4) y Draw the graphs of the 1 . y^ = %x. 3. >»=cos X. 5 . following functions : 2. y=sin x. 4. >‘=;c+8. 1*4. Classification of Functions (a) Algebraic and Transcendental functions function is one which consists of a finite number of SIX operations of algebra, v/z., addition, subtraction division, involution and evolution. ’ An algebraic terms involving multiplication. 4 A TEXT BOOK OF ELEMENTARY CALCULUS 6.X+2 Thus .\^H“5.v-r 1, (3.v t- 8.v-r 2)(3x 1), ^a_|_ ^ » (2x4-3)^ etc., are all algebraic functions. Any function which is not algebraic is called a Transcendfntal function. Transcendental functions arc further classified as under : (0 The Trigonometric functions such as cos x, cosec x, etc. (i/) The Inverse Trigonometric functions such as tan*^ x, sin’^ x, etc. (m) The Logarithmic functions such as loga x, loga (l+x), etc. (iv) The Exponential functions such as a'^, etc. (v) The Incommensurable powers of a variable such as x v 2, x\/3, etc. {b) Explicit and Implicit functions. A variable y is said to be an exp/Zc/V/MWC/ion of another variable .X, when, in an equation, expressing the functional relation, y occurs singly and alone on the left hand-side of the equation, and does not occur on the right-hand side. Thus in the equations v = 3.v“-r 5x4-1, — sin x-rcos x y is an explicit function of x. When y occurs mixed up with .x on one or both sides of the equation, v is said to be an implicit function of x. Thus in the equation -x^-i- y^=3axy y is an implicit function of x. (c) Single-valued and Many valued functions, y is said to be a single-valued function of x. if corresponding to one value of x, there exists one and only one value of y as in y— 3x4-4 etc. y is said to be a many-valued function of .x, if corresponding to one value of x, there exist more than one values of y, as in y-=4x, y*=8.x, y=sin”' x, etc. (d) Even and Odd Functions. A function /(x) is said to be an even function ol .x, if /(— .x)=/(x). For example /(x)=cos x is an even function of .x, for /(— x)=cos (— .x)=cos .x=/(x). It is said to be an odd function of x, if /(— .x)= — /(x). For example /(.x) = sin .v is an odd function of x, for /t— .Y) = sin ( — .v) = — sin x = —f{x). (e) Periodic functions. If /(.v)=/(x4-o) where “a” is a constant, then/(.x) is said tobc apc’r/ocy/c//mc/jo/i of X and “a” is called the period. CONCEPTION OF A FUNCTION 5 For instance, in /(a*)— tan x, /(n+x) = tan (Tr-rA') = tan x=f{x). tan X is a periodic function with ^ as the period. (/) Continuous and Discontinuous Functions. A function is said to be continuous or discontinuous according as its graph is continuous or discontinuous (without gap or break). For instance, the function >'=sin x is continuous, for its graph is continuous. This graph is given below : Similarly, the function j=cos x is also continuous. The function >»=tan x, is however, discontinuous because its 7C graph is discontinuous. There is a break at x — ^ and also at x= — The graph is given boiow : 6 A TEXT BOOK OF ELEMENTARY CALCULUS II 1. Define a function. State various kinds of functions, giving examples in each case. 2. lf/(x)=.x*-f 2-v — 5, find/(2),/(— 1),/(0), 3. If/(x) = sm -x- find/^Y 4 If Ax) - tan -Y, show that J\4 5. lf/(x) = log., X, show that f (nw)=f(m)4-f(n). 6. If = show that ^(tan 0)=cos 20. 7. If /(x) = x®, find {fia-\-h)~f{a)}/h. 8. lf/fx)=tanx, show that 2 tan h h /i(l— tan/i) 9. Prove that cos x+sec x is an even function. 10. Show that/(x) = 5x*+7.r*+x is an odd function. 11. If v=/C-y) = show that x^/O*). 4.V — 3 12 lf/(.Y) = 2xVl--x®, show that sin “ ^ = sin x. 13. If /(x)=| 7 ^}. prove that 14. If + = show tha^ , rv'±Y 4 I x=log|_ ^ J- 15. For what values of .v. have the following functions no definite value : (/) 1 x-r (i7) I (m) ~~ ? sin .v .y*-5.y+6' 16. Show that tan .v, sec x, cosec x, cot x are periodic func- tions. What are their periods ? CONCEPTION OF A FUNCTION 7 15. The phrase “x tend? to a”. Let A- pass successfully through an infinity of values according to some law such that for each value taken by .v, we can distinguish the values which precede from those which follow, and that no value of .X is the last. If now the successive values of x approach a definite number *a” in such a way that the numerical value af x — a becomes and remains smaller than every given positive number, however small, then we say that X tends to "a" {written as x ~*^a or x has the limit “fl”). It should be clearly understood that the number “o” itself may or may not be a value of a-, [f x takes on successively the sequence of values 1. 4", > then Lt a=0 even though 0 is not a value of v. 1'6. The phrase “x tends to infinity'’ Let .Y take a series of values successively which ultimately become and remain greater than every real positive number, how- ever great, then we say that '‘x tends to plus infinity" and we write it as x-> + oo. On the other hand, if the successive values assumed by x become and remain smaller than every negative number, then we say that "x tends to minus infinity" and we write x^ — oo, T7. Limit of a function Let us now consider the behaviour of a function /(. y) when x approaches any given value "a". As .y approaches the value "a", the value of /(x) may become closer and closer to a definite number /. Let us consider, for example, the function /■{.y)--2v4 I, and let x approach the value 1. Then it is easy to see that as .y becomes closer and closer to 1. 2x4-1 becomes closer and closer to 3. Further, the difference between 2x J1 and 3. /.<>. 2(x-l) can be made as small as we please by taking .y sufficiently close to 1. In such a case we say that 2x-l-l has a limit 3 as .y tends to L and write Lt Ax) -3. 1'8. Difference between the “value” and the “limit” of a function Let us consider the function For x=2. Ax)- 3, and, in general, the value of this function for x = fl is 8 A TEXT BOOK OF ELEMENTARY CALCULUS If, however, we try to find the value of the function for .v— 1, we face some difficulty. For l,/(.x)= o which is absurd. Thus, we cannot find the •ixact value of this function for x exactly etiual to 1. We can, however, find the limit of/(x) when x approaches 1. The behaviour of /(.\) when .v approaches 1 through values less than 1 or through values greater than 1, is indicated in the table given below : x= ■9 *99 ■999 * • * 1 • • • 1001 1*01 i ri /u)= 1-9 1*99 1-999 • • • ^ • • • 2*1 We notice that as x approaches 1 cither from left-hand side or f rom right-hand side,/{x> approaches 2. Hence ffie limit of/(x) for X approaching 1 is 2. In symbols, we write all this as Hence, we define the limit of a function as under : If a fuvetion fix) cppnaches a fixed quantity when x cpprcaihcs "a'\ then "I" is said to be the limit of fix) as x^a. In symbols, wc write Lt fix) = l. 19. Evaluation of the limit of a function The limiting value of a function fix) for a given value “o” of x may sometimes be the same as its exact value for x=a. Hence, in order to find Lt fix) we tirs.t simply substitute "a" for x and find A— ►U fid). This will give the required limit. But if fia) assumes a form which is meaningless, then wc first simplify fix) so that when we substitute "a" for x, it dees not assume a meaningless form, and in this way we get the required limit. This is illustrated below. Example 1. Find Lt ( v- * 2.V--4). -►I Sol. Here /(.v) = a--‘-2.\-*-4 Putting x = 2, weget /(2)-(2)"+2(2)4-4 12 which is not meaningless. Hence, Lt (.v3 • 2x • 4} -12. CONCEPTION OF A FUNCTION 9 Example 2. Find Lt x^2X-2 Sol. Let/(x)=^^ X ^ Putting x=2, we find that 1 ■which is meaningless and cannot, therefore, be the //>wi/ of/(v). 4 Hence, we try to simplify If -^=2, then je— 2=0 and we can- not divide .x^— 4 by .x— 2 for division by zero is not permissible. However, in the problem x is not given to be equal to 2. It approaches 2, so that x—l^O and we can perform the division. Thus, the given expression x“-4 (X-2KX + 2) X— 2” X— 2 If we now substitute 2 for x, x2-4 x-2 =x+2 takes the value 2+2=4 X*— 4 Lt ? x->2 X-2 = Lt (x+2)=4. x->2 Another Method (Most practicable) Let Then, Lt x=2+A 0 as x-»>2 x2-4 =* Lt {2+/»)2-4 x ^2 x-2 /“o (2+/i)-2 = Lt h-*0 " = Lt (4+6)=4. h -*0 Example 3. Evaluate Lt x-*.» 4x®— 2x+7 Sol. The symbol x-*oo indicates that x takes upvalues which go on continually increasing. Now, if x is large, both the numerator and the denominator of the given expression take the form which is meaningless. Hence, here also we have to sim plify the expression before we find the limit. Thus dividing the numerator and the denominator by x*, we have 10 A TEXT BOOK OF ELEMENTATY CALCULUS 5.x^-}-7.y-3 4.x^-2.x-^-7 5 + -^--^ .X .r* 4-^+4 .X .x® .V , 5.vM 7.x-3 _ 4x — 2X'^'7 2 7 4-----^ .X Now, as A- increases, that as A- and .X 1 •> • * • A- .. decrease so that we can say 0 Hence, the required limit— t- Alternative Method (Less practicable) 1 Let A* = Then y y-*0 as A-*co Lt .v-^ » 5 A- t-7.x — 3_ V 4.V-2X + 7 4 1,^7 r y - It 5 + 7>’-3/ "..o4-2y + 7y* - 5 . I'lO. Some Important Limits 1 I sin H 1. Ll -n-=l. ..-►0 ^ Let / AOP=*j radians. With O as centre and any radius draw a circle cutting OP, OA in P and A. Draw PM 1 OA and produce it to cut the circle in Q. Draw the tangent at P and pro- duce it to meet OA in T. Join TQ and OQ. Then the right-angled triangles OMP and OMQ are congruent. QA. MP=MQ and arc PA=arc Again, from the congruent triangles CONCEPTION OF A FUNCTION It or i.e. or or i.e. Now chord PQ-t: — > cos 0 lies between 1 and cos 6. But cos 0 ->l as ©-►O. approaches 1 as 0 approaches 0. Hence Lt — 5— = 1 * 0-^0 ^ 2. Lt 0-J.0 ® Sol. L.H.S.= Lt tan 0 o^o sin 0 n = Lt b O-^o COS0 Lt o-»0 0 = Lt cos 0 o-*.o =T = ‘- V**— Lt - — x-^a ^ ^ Solution. Let Lt Lt x->a ® h -^0 0 -{-h—a x=a-\-h so that h~*-Q as x~^a ^ {a+/i)"— a" = Lt A-^0 (-ir- fExpanding by Binomial Theorem] 12 A TEXT BOOK OF ELEMENTARY CALCULUS = Lt /«-»0 “•[ 1 +"i+=%" ]- Q^-Vna^-^h = Lt h-^-O h n(/i— 1) 2 ! a"-3A*+...— o’ = Lt r na’-‘+"-^^’a"-^A+ 1 h^O L ^ ’ J ==na"-^ Worked ont Examples Example 1. Evaluate Lt x->2 3 a*+.x— 14 0 Sol. The expression takes form if we substitute in it 2 for .Y. Therefore, we write 2x^-hx-\0 = Lt (.y-2)(2.x+5) x^2 3x»+.x-14 ^^^2 U-2)(3x+7) = x-*2 3.X+7 4 + 5 ^ "6 + 7""l3 ■ Or thus, Let x=2-rh, so that as .x-^2, h^O. I t 1 , 2(2 + /f)^+(2+A)-10 *• + /.io 3(2 + /))^ + (2+A)-14 /,^o3/i*-f 13/. It 2A+9 9 '■/.Xo 3/1+13“ 13' Example 2. Evaluate Lt + Sol. Lt ^-^4±+=-!= Lt 4x*-2x-r7 5+-^-+ X X* 4-^+’ X X* (Dividing the numerator and the denominator by x*) CONCEPTION OF A FUNCTION 13 Now, as x->co, and ^0. Hence the required Or thus, Let x=— , so that as .v -»■«>, y-*0 y Lt 5x»+7x— 3 _ f y 4x^— 2x+7 ^^0 4 / y __ j 5+7v-3r "^r>0 4-2>-+7/ ^_5_ 4 ■ Example 3. Evaluate Lt [ v'x-^+5x+4- v/x^-3x+4] Sol. Lt [v/x2+5.v+4- \/.>r-3A'+4J v/x2+5x+4+ v/x''-3x+4 (Please note this step) * Lt * = Lt (x* -i- 5x + 4) - ( X* - 3.x+^ = Lt v'x2+5x+4+Vx^-3x+4 8x = Lt Vx*+5x+4+ N/x*-3x-f 4 8 ^+1 (Dividing the numerator and denominator by x) -I- =4. Example 4. Evaluate (0 Lt X-* / sin ax "nA: 0 \sin bx ) (//) Lt x^O l—cos X '2 (i7/) Lt sin bx sin (x+A)— sin x h 14 A TEXT BOOK OF ELEMENTARY CALCULUS Sol. (0 Lt ')*=-. ;*:^0 \Sin bx J 'sin ax \ k ax Lt I — : j— x^O 1 sm bx bx Now Lt = Lt = 1 x->-0 ; r ->0 Required limit 07) Lt - v-^O 1— COS X 2 sin X = Lt x->0 = Lt .2 2 « 4 sm (6x)* (Please note this step) 1 jr->0 Sin X Now Lt . T ->0 “ Lt 2 = 1 The required Lt = Lt Jc)(l+cosx) x-^-O -XHI + COSX) = Lt (5iliL\ x->-0 \ X J X 1 j_ l+cos X 2 (i7i) Lt (x”t~A) sin x 2 cos ( x+ Lt Ar ->^0 ( ^ sin = Lt cos (x+|-)sm 2 (Please note this step) CONCEPTION OF A FUNCTION 15 h sin y Now Lt — = I and Lt cos /j->0 Ji_ /i->0 2 The required limit— cos a'. 1*11. Theorems on limits {The proofs of these theorems may be taken for granted at this stage) If Lt f{x)=l and Lt ^(x)=m, then x^Q x-^a (<) Lt [/W±^{.v)]= Lt f(.x)± Lt x-*‘a x-^a = t±m. (ii) Lt [f(x)t)]= Lt f(x)X Lt 4M = lm. x^a x^Q x^a ft 1 t W) (jv) Lt [cf{x)]=c Lt /(.x)— c/, where c is a constant. x-^a x-¥a (v) Lt log /■(x)=log Lt /(x) = Iog I provided />0. x-^a x-*a 1 . 3. 5. 7. 9. Exercise III Evaluate the following : Lt x-»-i xa-25 x-5 ' 2. Lt x-*-3 a in 00 T T ^ H 1 1 H H Lt x-*a X”* — fl"* x—a ' 4. Lt Vl+x-l . (K.U. 1976) Lt X'* • 3x2-4x-I5 5x2-9x-18* 6. Lt X-*- • 3x2+2x+l 2x*+3x+5 * Lt X~¥ <*> n(n+ 1) 8. Lt x-*0 vr+i-vi^ (n+2)(/i + 3r X Lt x^a X 10. Lt [ Vx-M— , ;c]. l-Vl-x [Hint : rationalize it.] »• „l;‘.(T+i+i+ 12. Lt + +ar"-*) ; I r ( < I. A TEXT BOOK OF ELEMENTARY CALCULUS 16 13. Lt (1®+ 2=*+3*4- +n»). x-^ » ^ tan X— sin x 14. Lt x^O x^ 15. Lt sra mx 16 Lt (x+A)— tan x x-*.0 A sin 3x cos 2x cosec x—cot X 18. Lt x^O 20. Lt O-^O sin 2x tan 0— sin 6 17. Lt x-»-0 19. Lt 0->:: 2 cot 6— COS 0 cos* 0 0* Show that : 21. Lt = — o'" " when m>n. x^a 22. Lt . v -»-0 23. Lt Y-^-O X- — n cos q.v— cos bx _ b-~-Q“ X* 2 Vl+X+X*— 1 _ 1 2 X 24. Lt U+/;)-sec x _ A^O ^ sec X tan x. 25. If /(.y)=x*, show that Lt /ll ±^)-/< v) *-►0 n . v ”'— 1 26. Lt . x-^l ^ * 27. Lt (sec .X — tan .v). X~*n',2 {K.V. 1975) 1 28. Lt X sin 1 29. Lt /i>»0 5+h 5 30. Lt x-^0 {K.U. 1975) VT+x*-‘/f+^ {K.U. 1976} iK.U. 1975} {K.U. 1975) {K.U. ]97S> M 2 Differentiation 21. If a variable x changes from one value Xj to another value X 2 ,* then the difference between the two values is caJled the increment in x, and is usually denoted by 8x (read as delta x ). 2*2. Differential Coefficient Def. The differential coefficient of a function is defined as follows : Ut/(x) be any function of X and /(x+5x) the same function . . . Ax+Sx)— /(x) .. i... of x+8x. then the limit value of the expression — as 6x approaches zero is called the Differential Coefficient (or the Deri- vative) of /(x) with respect to x and is denoted by / (x). 2*3 Another form of the definition of the Differential Coefficient Let v=/(x) be a given function of x. If we give to x a small increment Sx and the corresponding small increment to y, then ihc limiting value of the ratio “ as8x->0 is cB\\ed ihc Differential coefficient (or the Derivative) of fix) or of >- and is generally written as ^[/(x)] or ^ or /'(x), etc. Thus if We have y==f(x) y-\-^y~f(x-^Bx) S>'=/(x+Sx)-/(x) Sy /(x+Jix)— /(x ) ^x ^'x dx Lt 8 . v -»0 oy Sx = Lt 8 a ' -^0 /(x-h^x)-/(x) d'x ^ is read as “dee-wy-over-dcc-eks”. dx Note 1. 18 A TEXT BOOK OF ELEMENTARY CALCULUS Note 2. In the ratio S.v and Sy being small increments in X and V respectively have meaning though taken separately as well, while cix and dy in ^ have no meaning when taken separately. On j the contrary, ^ is a smg/e quantity standing for 2 4. Working Knle f-r finding the differential coeffiaent of a given function f(x) 1. Put V*/(.x). .. 2. Give to -v a smaU increment 8x. and the corresponding small increment By to the dependent variable y. Thus if y^fix) we have y+Sy=/(A'+Sx) 3. Subtract (/) from (ii). then ?iy=fix-\-Bx)~fix) Divide both sides of (iii) by Bx. Proceed to the limits when 5x-*0 ...(I) ...(lO 4. 5. and denote Lt ^ j . dy Note. 1. From {Hi) it is easy to see that as 8x-*0. in the manner Note 2. Differentiating a given Th/arw-n func- explained in Article 2-4 as known as d^crenliatang the given fun tion ab-initio or from first principles or from definition. dy 2 5. Geometrical Significance of ^ Let P(A, y) be any point on the graph of the function y fix) -and Q (A-f Sx, be the neighbouring point on it. Draw PL and QM perpendiculars on OX and cular on QM. Let the secant PQ on being produced meet x-axis m N making an angle 8 with its positive direction. Now PR=LM==OM-OL=(x+Sa)-a=8a and QR=QM-MR=QM-PL=(y+Sy)->'=Sy. From the right /_d APQR. we have - *x =tan 8. Now, as Q begins to move towards P ^Jong the cu^c, S.v decreases, and when Q hnally coincides with P. 3x becomes mfinitely DIFFERENTIATION 19 small and approaches zero. Consequently, the secant PQ becomes tangent to the curve at P and the angle 0 is changed to angle >1', T O N L M which this tangent makes with the positive direction of x-axis. This means that Lt 1-t tan 0 —tan 4'. 8x-f0 Q-^p But from the definition of differential coefficient (Article 2’3), we know that T t 8x^0 Sx “dx dv 4'“slope of the tangent. Hence, the value of the differential coefficient of a function f{x) at a point is the slope of the tangent to the graph of the function at that point. 2*6. Differential Coefficient of (0 *" (//) (ax+b)". (0 Let >'=x’‘ Then >»+S>;=(x-i- Sx)" S>- = (x+Sx)»-;» =(x+5x)'‘— X" . r/ . . sx x" 1 Now 1 + )■-] involving third and higher powers of 20 A TEXT BOOK OF ELEMENTARY CALCULUS ] 1 5.x “ n_i , n(/i— * ' 1 • =/i.x*’ Sx + terms mvolviog second and higher powers of Bx. Hence. ^ ~ dx 8.V-+0 , . r , «(n— l)x"*2 , “1 = Lt 1 n.v” *H =-{ .5x+ 8.t-^0 L 2 ■ -I = «.x""'. 07) Let v=(flx+&)'' •■•(1) -f Sy = [a(jf 4* S.x) -f W = [fl.x4 b+aKx]’' -<-»■ [ ] = tflA-r i7)"~f n(q.x + 6)"~^oS.x+ — (ax4-/>)"~^ a’(5x)®-i--” •••( 2 ) Subtracting (1) from (2), we have Sx+^^^-j^ (ax + 6)"”®fl*(5x)2+ Dividing by 5x, ^ =n (ox+*)"-La + -^^^^Vax4-6)'-V-.Sx+ = T t ^ dx 8^_^o = Lt rn(ox+A)-”ia4-^^^^^^(ax+W'-*.fl*.Sx 8.x-»0 L ^ » 4- terms involving second and higher"] powers of ^x J = na(ax-\ b)”-'. DIFFERENTIATION 21 2-7. or Fundamental Theorems of Differentiation Theorem I. The differential coefficient of a constant is zero. Proof. Let y=c, where c is a constant (No increment is given to a constant) ^=c— c=0 Sx dy Hence ^ = Lt ^=0. Theorem II. An additive constant disappears in differentia- tion Proof. Let y=f{x)-rc y S^= f{x '”8x) -r c h‘=f(x + ?ix)~fix) ?iy f(x+Kx)—f(x) fix d'x Lt Lt dx Sa'->0 ^X Sy ^0 Thus “c" has diappeared from the process. Theorem III. The differential coefficient of the product of a constant and a function is equal to the product of the constant and the differential coefficient of the function. Proof. Hence Let >’=c/(x) Zy=^cfix + ?ix)-cf{x) =c[/(xfSx)-/(x)] ^ fix±fxy^jXxi Sx ^ Sx L. ^ = c Lt . A^+ |£ )-/(x). Theorem IV. The differential coefficient of the algebraic sum of several functions is equal to the algebraic sum of their differential coefficients. Proof. Let y=u ^v—w-\- •••(!) when u, v, are functions of x and y is also a function of x. Given an increment Sx to x, then as u, v, w, , y arc func- tions of X, the reactive increments S«, 8v, S»v, , Sy respectively. Then ;»4-S>'=(w+S«)4-(v4-Sv)— (^+8^) + ...(2) 22 A TEXT BOOK OF ELEMENTARY CALCULUS or Subtracting (1) from (2), we have — SwH- Sy Sm , ^ 8.\~~Bx 8 a 8 a • Lt 1^- Lt dx * x -»0 Sa --*-0 ~dx '^dx dx Hence P-+ Lt 5a 5^->o ?L 8a - Lt 8w 8a Exaxnple 1 . (0 V-V- Solved Examples Find ab-iniliOy the differential coefficient of {K.U. 1975) (») Sol. Let y=y/x then y+Sy=V A^8A 5y=V^A-h Sa— V'X 8v VxT^x—y/x ♦ 8a 8a _ Sa 8a(V a+8a+ V x) 1 V A 8a 4- V -Y (Please note this step) Hence - Lt S\-*.0 I L-. Va-tSa-I-Vy -Vy (/;) Let then v+Sv= I 8> (.Y-' '’' v)-r o 1 1 f-.— ( V" 8a) + o A+fl ( A -i- fl) — Uy h 8 a) -f oj (A-ra)[(A-r8A)^o] DIFTHRCNTIATION 23 then or or — Sx Hence (x-l- a)l(x+6x)-i-fl] J[ dx dy (x^- a)[( X 4- 5x) + a] ^ Lt ;:^=- 1 dx s;::;o ix-\-ay Example 2. If ;»=Ax®+Bx2+Cx4-D, prove from first principles =3Ax“+2Bx+C. Sol. ;»=Ax®+Bx= + Cx+D _y-l-S^=A(x + Sx)=*4-B(x+Sx)a4-C(x + Sx) + D y+ By= Ax^ -f 3 Ax^Sx + 3Ax(5x)2 + A(Xx)^ 4- Bx* + 2Bx.ax 4- B(ax)^» 4-Cx4- CSx 4- D S>»=3AxaSx4-3Ax(5x)34-AC5x)®4-2Bx.Sx 4-B(Sx)a4-CSx :=3Ax2-f3Ax.Sx4-A(Sx)24-2Bx4-B(Sx)4-C Lt Lt [3Ax^-4'3Ax.Sx4- dx 8^_^0 8x^0 A(Sx)«4-2Bx4-B(Sx)-}-Cl Hence = 3Ax*4-2Bx4-C. 1 . Exercise IV Write down the derivatives of the following : 1 (i) X*. Hi) 1 (Hi) Vx. (iv) X®. (v) ^ (v/) (2x1- 3)1'’. (v/7) (8-13x)*'^ (v/7i) 1 2. V 5x4-7 Find the differential coefficients of : (i) x(l-x). (//■) x3-6xH-llx-6. (in) x^4-5x*-7x24-4xH-5. (iv) ~ t (v) .r*4-3>4H-4 ( VI ) mx 4 a 3. x* * m (vi7) 7x'*4-8x-\ (0 If ^(x)=px”*4-9x’*, find ^'(x). (i7) If/(x) = x*-l-4x^4-6x*4-4x4-l, find /'(x). (///) If;;=l~2r i 3l*-4/3, find at 24 A TEXT BOOK OF ELEMENTARY CALCULUS (iv) Find ^ atx=2. if >'^.x^+3,x5 + 3x+ 1. {iv (v) If >»=x*+5x+6, find ^ alx=3. 4. Find ab-initiOy the derivatives of : (/) 5x*. (ii) (3x+4)3. {Hi) (iv) ax^-\-bx+c. (V) (ix) (viO 3x+l 5X-I-7' (v/fi) x+1 fxl 1 (xi) \/X ‘ V ax -f b' x*+3x9 x+3 -18 1 • » ixiii) 1 1 3x + l‘ v' flX-i ax+b iK.a.:97n 5. If A denotes the area of a circle whose radius is r, show that ^ is equal to the circumference of the circle. [Hint. Area=nr-]. dv 6. If V is the volume of a sphere with radius r, show that jp is equal to the circumference of the sphere. 2 8 Differential coefficient of the product of two functions Theorem V. The Jijferential coefficient of the product of two funcliom i: equal to the iirst fimctiony. the differential coefficient of the second fwiaion-stcond function X the differential coefficient of the first function. Or symbolically iL = u — -V where u and v are both functions of x. dx ^ ^ d.\ ' d.\ Proof. Let v=mv and let <5.v, o//, Sv and Sr be the incre- ments of .V. H, V and y respectively, then Sv== (m4-^«1(v-1-Sv) % • * = nv-! t Sm.Sv— wv = ifh' f or s': Proceeding to the limits as S.x-+0 and consequently Su, Sv, Sy-+0. DIFFERENTIATION 25 and u dy dv , du Hence, /“ = w:j-+v^- dx dx dx Cor. If y=uvw, where u, v, w are all functions of jc, then dy dw , du , dv _=m. — 1-^^, dx ^ dx dx Proof. Let z=uv, so that y=zw By the above theorem, we have dx dy dw dx di 1 -\-W j~ dx ■..(1) dz dv du dx ~ “ dx ~ dx -(2) dx value of dz dx from (2) dw , ( dv , ^ dx !-w[ u — f-v dx dw ~\-uw dv , = uv dx du dx du dx ) • ■•(0 J’=MVW + Sy = (u 5 w) ( V H - 3 v) ( IV 4- S w) Substituting, we have ^y=(u-\- 3u)(v + 3 v)(h'+ S w) — uw 8>> = w viv-f )tv5m+ uw'8v + wvd >v-|- v8 w vSh-Sm + H'5m3v4'8w3v5iv — uvyv Sw , 5v 3>v Sv 8x ax -^v a.T ■ .3v'.avv Let Sx->0, 8u, 3v, ew, 3;; alI-^0 dy du -*-=uu» _ dx I dv dw ^ ^ coefficient of the quotient of two Co-efficien, of ,hc gucien, of ^(D eno.)(Diff. Co-effi. of Numer.)-(Numer.)(Difr. Co-effi. ot Deno.) Square of Denominator 26 A TEXT BOOK OF ELEMENTARY CALCULUS or symbolically of X. when u and v are functions Proof Let y=“ and 8.x, and Sv be the increments of x, V u and V respectively, and 8y the increment ofy . t t then y+Sy = 8y= v+ 8 v v+ 5 v V mv+v. 8 m — m' — u. 8 v v(v+ 5 v) vCy-f-Sv) Now as 8.V Hence Su Sv By ^ ~ Sx ~ v(v-i-5i’) 0, 8v-^0. (v+8v)^v du . dv dv ^ dx “ dx " v» Example 1. Sol. Here Example 2. Solved Examples if y=(2.V'^ 3)(4.x4 5), find^- y=(2A-r3)(4.x4-5) ^=(2A-h3) ■j^(4A+5)-)-(4.V'f5)^ (2.v-f3) = (2a-^3)x4 + (4x^5)x2 = l6.v-r 22. Find ify=U"4-I)(3A^4-9X2A-i-4). ax Sol. ^ = {x^+\)(Xx^-^9)4- ^ (.x*+l)(2A--i-4) dx dx d dx Ua^ + 9) + (3a»+9K2x ' 4)X^ (a* 4 - 1 ) = (a- + 1 )( 3.>4 + 9 )x 2 +(a 2 -|- l)( 2 .x-i- 4 )x 9 x* 4 -( 3 .x*+ 9 )( 2 a 4 - 4 )x 2 x = 2(a'^+ 1 )(3.v^ + 9) -r 9 a*(a2 4- 1 )( 2x4- 4) 1 2.x(3.x-* 4- 9) X ( 2 .x-!- 4 ). DIFFERENTIATION 27 Example 3. If find dy dx Sol. dx'' (x^+xy 2x(xa + l)-2x(x^-l) " (x^+l)" 4x “ (;c*+l)*‘ Example 4. Use — form to show that V Sol, Here u=i and v=x d t i \ xxO-l.l r d ^ ^ ^[ir) — — L-' 1 Note, Questions on uv and ^ forms from the forthcoming exercise should not be attempted till the following theorem is fully followed. 2’10. Function of a Function Sometimes it so happened that is not defined directly as a function of x. On the contrary, a new variable, z (say) is introduced, and y is defined as a function of z, which, in turn, is defined as a function of x. This means that is a function of z while z is itself a function of x. In such a case, y is said to be a function of a function. For Example, if y—w" ...(/) and u=(x*+3x2+5x+l) ---(/i) we say that y is a function of a function . If we eliminate u between {Oand(«)» we can get a direct relation between y and x, viz., y=(x*+3x*+5x+l)'' from this relation, we can got ^ direct. But, sometimes this sort of elimination becomes very difficult. In such a case, we make use of the following theorem for finding 4y-, dx 28 A TEXT BOOK OF ELEMENTARY CALCULUS Theorem VII, Given (i) y —f (z) and («) z=^U), to prove that dy dy dz Proof. Let and S>’ be the increments of x, z and y respectively, such that (а) z becomes z+Sz as y becomes y-\-^y in (/), and (б) X becomes x+Sx, as z becomes z+8z in (ii). Now, by ordinary Algebra, ^ = 1^ however small $x^ or Sz may be. Thus, in the limit, we have Sv . - Sz Lt ^ — Lt T - X Lt S v-*-0 ix-*0 6 x-¥0 dx or dy dz ^ X dz ''dx Cor. Differential co-efficient of u", where u is a function of x, n" is a function of u which, in turn, is a function of x. Note. The differentia! coefficient of m" with respect to « is while that with respect to a- is /)«"■* Similarly, the diffe- rential coefficient of a" w.r.l. a is /ja""*, whereas that w.r.t. >’ is dx nx dv Thus, it is V ery important to know with respect to what we are required to differen tiatc a certain function. When nothing is mentioned in this behalf we can presume that the function is to be differentiated w.r.t. its argument. This means that if we are required to differentiate a", we have to differentiate it with respect to its argument which is x. Its diffe- rential co-efficient will, therefore, be «A""h But, if we arc required to differentiate a" with respect to t, the differentia! co-efficient will be .n-l dt Extension. The result of Theorem Vll can be extended. If V is a function of u, u a function of v, and v a function of x, then nx DIFFERENTIATION 29 Theorem YIH. Relation between ^ and ~ • dx dy If is a function of X, then we can express x In terms of v Hence x can be regarded as a function of>;. From the former relation we can calculate ^ and from the latter — • ax dy They are connected by the relation dx = 1 . dy dx dy Proof. By ordinary Algebra, we have Sx When Sx-^0, Taking limits, we get dx dy Cor. ~ and ^ are reciprocals of each other. Sot. so that Example 1. Let Solved Examples Find if>' = (.x<+7.v- «=x^ + 7x2+3 3)®. y=u^ Now from (i). du =4x^4- 14.x: and from (ii), Hence dx du = (4x3+I4x) dx du dx •••(0 -.(/o =8(x^+7x»+3)’ (4x®+14x) = 16x(2x2-|-7)(x*+7x2^-3)^ Alternative Method. The given expression (x^4'7x*+3) function of x^4-7x*+3 which is, in turn, a function of x. d e IS a dx 7x^+3)* = 8(x<+7x* + 3)7X ^^(x<+7x“+3) = 8(x<+7x^ + 3)(4xH-14x) = 16x(2x-+7)(x*+7x2+3). 30 A TEXT BOOK OF ELEMENTARY CALCULUS Note The student is advised to completely master the alter- method, as he can save a lot of time by making frequent use of it. y r • Example 2. Show by the method of function of a function that the differential coefficient of (flx+6)" is na(ax-\-b) Sol. Let y=(ax + br and u-iax-\-b) so that y = a’* From (/) and from Thus ax dx du dx = nu^~^y.a=aniax-^by*-'^ . . (After replacing a) Alternatively Example 3. Sol. Let + ^(ax+b) £ dx ■ ' dx Differentiate Vx+fl — y/x—a .X+fl+ ^ X— <2 \/x+Q— y/x—a ( j2 %/ x+^— V x—a V '' form s ) Rationalizing the denominator, we get __ [ V (x+g)— \^x— 0)3* x-Fo+.T— fl— 2 V/ X*— fl* " 2o a DIFFERENTIATION 31 Note, in example (3), Vx-- + io). 14. 15. (x-1)2(4a + 7). 16. 17. 18. Differentiate the following : 19. bx’-\-c x^+a 20. 21. fx^-2ax . V bx-cx* 22. 2ix-\) 23. x^+2x~5 24. Vx+i+ V x—1 ‘ (3Aa+4)*(l-x^)®. {KV. 1977) iP.U. 1952) V q+x— y/a—x V a-\'X‘^ V a— X* 2*11, Implicit and Explicit Functions Def. “Jf the relation between x and y be solved for in terms of A in the form of the simple equation y=^Ax) 32 A TEXT BOOK OF ELEMENTARY CALCULUS then V is said to be an Explicit function of If the relation betwwn X and y is not expressed in this form, then y is said to be an Implicit function of x.** 2' 12. Differentiation ofy" with respect to x Symbolically, Let then d dx Now, by the Function of a Function method du dx dy dx and from (0» Hence 1 dx The student is advised to note this carefully. The differential dv coefficient of y^ with respect to x, as shown above, is ^ and not u.ilv L On the other hand, hv"-^ is the differentia! coefficient of y" when the latter is differentiated with respect to y. Example. 3v2 is the differcnlial coefficient of / with ' dx respect to x, whereas 3y" is the differential coefficient of with respect to y itself. 2' 13. Differentiator of an implicit function When y is an implicit function of x, it is mixed up with x in such a way that at times it is difficult for us to express y in terms of X or X in terms ofy. Consequently, we cannot easily find or In such a case, the following rule is laid down : (1) Differentiate both sidc> of the equation with respect to x. (2) Solve the resulting equation Example. If fl.\■^+2/;.vl’ + ^y®= 1, find dy dx DIFFERENTIATION 33 Sol. Differentiating both sides of the equation w.r.t. x, we have or 4(^)+2/,^ (x>.) + ir^(/) = f (5) or a|^(x»)+2A( y ^ (Please note this step) or ax2x+Vi^yxl\-xxlx^^+2y^Xb=0 or or or (/ix+M ^ = — (ax+Ay) Hence dy __ ox+Ay 5x Ax+A;^ ax^ H- 2Ax>' -f Ay* — 1 giving increments a(xH-Sx)* + 2A(x4-Sx)(y+Sy)+A(yH-8y)*=l ...(2) Subtracting (2) from (1), we have a(2xSx+ Sx^J + 2 A(y 5x +xSy + SxSy) + A(2y Sy + 8y*) = 0 Dividing by 8x a(2x+Sx)+2A [ y+x^+Sy ]h-A ^y ]=0 Let Sx-^0, 8y also ->0 fl ( 2 x ) + 2 A [^+ x ^] + 26>.^=0 d^ (gx+Ay) dx “ (Ax+Ay) 214. Diflferentiation of functions expressed in parametric forms Sometimes y is not expressed as a function of x. On the other hand, both x and y are expressed as the functions of a third variable. 34 A TEXT BOOK OF ELEMENTARY CALCULUS For instance, these are expressed as the functions of by means of two different equations as shown below : .x=/(0 ; y=^{i). In such a case, we have to eliminate **/*’ from the two equations before differentiating y w.r.t. .v. But it is not always easy to effect such elimination, and obtain a direct relation between x and y. In such a case, ^ is obtained by the Function of Function Method. From the equation x=f{t) we get as /'(^)* and from the equation >>=^(0 we get ^ as 4>'{t). Finally “ is obtained as under : ^ ^ j dx f dx dt I dt ~~ dx it) it) The method is illustrated below : dv Example. Ar=o/®, >’=2or, find dx Sol. From x=at^y -j; = 2at at dy and from y==2at, ~^ = 2a dx dt dx ' 2at t Solved Examples Example 1. If ;c”’4-y"=n'", find Sol. Differentiating both sides w.r.t. x, we have mx m-l dx 4y dx .m-l dy Example 2. Find ^ when : 2t l-H/ a* y= i-f* 1+/* DIFFERENTIATION 35 2 . 3, 4 . Sol. x= It ^ _ 2(1+ f «)-2fX2/ dt (1+^*)" 2(1 (l+/‘)* 4r ^=TT? ^/v -2f(l+r*)-2f(l-/*) " (l+r*)* —4/ “ ( 1 -/*)= ( 1 +/*)* 2 / _ Hence ^ X J -(l + r‘)*^2(l-/*)* r»-l « with respect to x* 1+x Example 3. Differentiate Sol. Here we have to find Af_£!_V V » I Z— X* Now >>= l+x» dy 2 x(l+x»)- 2 xxx* dx (l + x»)» dx = 2x 2x =(T+?V‘ from (1) and (2), we get 4y —^y — dz " ’ ^ " 2x 1 5;^^dF'a+x*)*^2x 1 (l+x“)*’ Exercise VI Find ^ when : dx 1. x= 2/ x= 1 + /* ’ ^ ' fld-/*) 1 + r* » y 1+r* x=a/*, y~2at. x^ct, y=~ l-f» i+f* 2bt - 3»+c=0. 7. a' 9. .x^+,v*=3x>'. It. 1 3 . fl.x* + 2hxy + by^ 2g.x + Ify 4- c = 0. Diflferentiate ; 14. 15. 16. 17. 18. 19. 8 . 10. ax^-\-2Jixy-^by^^\. 12 . xy=<^. (K.U. Nov. I97S) x" with respect to x*. 7x*— 12x3 7x*— 15x. .X w.r.t. X*. 14-.X* X— V l^x* w.r.t. V 1— X*. ax — 6 . a'x+b’ cx-^rf cx-\~d Miscellaneous equations : If .V— \/.x-hVx— ^ 'X+v'X- to oo show that (2_>’ — 1) ^ = 1. (K.U. 1977) 20. If ;-=X'^ 1 x + 1 x— 1 X + . to prove that 21. Assuming that 1 -hx+x*+.x*+ dx 2.V— X +x"= 1-X-+1 1— X * find the sum of the series H-2x+3x* + 4-mx""' by differentiation. 22. Assuming that (l+x)" = H--CiX+''C^*+ +-CBX" show that "Ci+ 2"C3+3''C,+ +n"C«=n.2"-». [Hint. Differentiate the equation w.r.t. x and put x— 1.] X 23. If y= V 1— X , prove that ‘ - T y- Differentiation of Circular and Inverse Circular Functions 3-1. In order to find the Differential Coefficients of circular and inverse circular functions, we have to make frequent use of the following important theorem on limits. This theorem has been proved in Article I'JO. Lt 0-*-O sin 6 where 6 is measured in radians. 3*2. Differentiation of Circular Functions (i) Differential Coefficient of sin x. >'=sin X Then, ^-rS>’=sin (x+^.v) 5;’=sin (x+Sjc)— sin X \ . Sx T j T / sin P— sin Q=2 cos sin =2 cos I JC + ( P-Q 2 (Please note this step) It ^ ax-*.o 38 BOOK OF ELEMENTARY CALCULUS A TEXT Now Lt and Bx ^ Lt cos(jc+^)x Lt V 2 y 8x-* sin Sx-^-O cos ^ ^+"2" )■” cos X sm 8x 8 x -»-0 ^ 2 (Proved above) Hence dy =cos X dx -f- (sin ;c)=cos x. dx (lO Differential Coefficient of cos x. Let >>=cos X. Then y+ 8 y=cos (x+Sx) • Sv=cos (x+ 8 x) cos X • • » or = —2 sin ^ x+ 5x ) T \ = _2sinP^sin?^ COS P— COS Q = —2 sin Sin 221 — Sx -2 sin ( sin Sx 2 Sx . 8 x . ( , Sx = -sm^^x-r-2 j 8 x 2 (Please note this step) Lt ^ 0 -Sin ( -^+y) sin Sx S^. 2 DIFFERENTIATION OF CIRCULAR FUNCTIONS 39 and = Lt [ -sin ( )]x,Lt^ Sin 8x Sx Now Lt hx -*’0 [ -sin ( A+"^)]= -si Sin X sin 8a Lt ix ~*0 -Sa = 1 (ix =— sm A Hence dx (cos a) = — sin A. (///) Differential coefficient of tan x. Let >>=tan a Then >'+S^=tan (a+Sa). 8>>=tan (a+8a) — tan A sin (a+Sa) sin a ^ ... ^ v = : — r-jr-: (Please note this Step) cos (A^-8A) cos A sin (a-I-Sa) cos a —cos (a+Sa) sin a “ cos (a4- 8a) cos A _ sin (a+8a — a) “cos (a+8a) cos A sin A cos B— cos A sin B = sin (A— B)] __ sin Sa “cosT-’cH'Sa) cos A Sy 1 sin 8 a " Sa ~cos {a+8a) cos A* 8 a Lt dx ix-pO = Lt I X Lt sin Sa ix-p-0 cos (jc+8a) cos A Sx ^ , sin A But Lt ix-pO 40 A TEXT BOOK OF ELEMENTARY CALCULUS and 1 I siio cos (x + 5,v) cos .V cos® X dy 1 dx cos® X = sec’‘ -V. — (tan x) = sec® x. Hence Alternative method (not by first principles) Let y=tan x= sin X cos X 4 y. dx cos X . ^ (sin x)— sin x . ^ (cos x) (cos x)® c os X . cos X — sin x (—sin x) cos* X cos* x + sin* X 1 Hence cos* X d cos* X sec* X. dx (tan x) = sec* x. then (iv) Differential coefficient of cot x. Let y=cot x y+Sv=cot (x-i-Sx) Sy=cot (x •-S.v) — cot X sin X cos (X'i-Sx) cos x sin (x-f Sx) sin (x--5x) sin x) sin f — ^x) Sj; Zx sin ( A' x + ^x) sin (X - -^x) sin a ~ sin (x + 5x) sin x — sin‘'^x r.. ^ ,,, — ; — , s> ■ ■ — [ . sin ( — 9) = sm (x-!-ftx) sin X sin Sa' 1 or Lt s dx 8x->0 Hx ' sin (x-r5x) sin x , . / sin Y \ , ^ 1 = Lt I —“ 5 — )x Lt — - — - 8,-^o ' ^ St-*.osin(x+ + Sx) sin X = -l X 1 sin* X sin* X = — cosec- X 4 ~ (cot x) — — cosec* X. dx sin 9 Hence DIFFERENTIATION OF CIRCULAR FUNCTIONS 41 Alternative Method (not by ab-initio) cos X Let ;/=cot x= dy dx sin X (cos x)— cos xx^ (sin x) sin^ sin y ( — sin x)— cos x (cos x) then sin^ X _ sin^ x+cos° X sin* X ~ = —cosec* X. (v) Differential coefficient of sec x. Let y=sec x >;4-3>'=sec (x-r3x) Sy=scc (x+8x)— sec x 1 1 1 sin* X or ‘cos(x-}-8x) cos X (Please note this step) cos X— cos (x-l Sx) cos (x+8x) cos X Sx 2 sin in( X+y 1 Sin y COS (x + 8x) cos X [^V COS P-cos Q = 2 Sin^^ sin in f 2 sin x+ 8x ) 8x COS (x+Sx) COS X 8x sin 2 8x sin ( x4- ) 5x sin 8x cos(x-h8x) COS X 2 Lt ^ dx dx = Lt sin f x4- 8x ) »,^0 cos(x~t-8x) COS x^ 8x Sin Y ~Jx^ 2 A TEXT BOOK OF ELEMENTARY CALCULUS sin X . _ siTi x i _ ”cos*x^ cos X cosx =sec X tan x Hence dx (sec x)=sec x tan x. Alternative Method (not by ab-initio) Let >=sec cosxX^ (l)-lx^ (cosx) cos® X cos XXO— 1 X(— sin x) sin x ^ Eo?x cos* X sin X 1 _ _ _ ^ sm X cos X cos X (vi) Differential coefficient of cosec x. COSCC X =scc X tan x. {K,U. Nov., J975) then ^+S^s=cosec (x+^x) Sy^cosec (xH-Sx)— cosec x 1 1 “sin (x+8x) sin x (Please note this step) sin X — sin (x+8x ) sin (x+Sx) sin x / , Sx \ / X— x+8x \ 2 cos^ x-l- y j sin ^ 2 / = sin (x+Sx) sin x 2cos( sin(-^^) ~ sin (x+ Sx) sin x 2cos(x+^J . , Sx\ ^ x( — sm ) in X \ 2 / sin (x+Sx)sin cos ( -f ) [V sin (—6)=— sin 9 sin(x4- 8x)sin Sx X - . Sx sm y DIFFERENTIATION OF CIRCULAR FUNCTIONS 43 Hence Lt Sx-^O cos = Lt ( x+ Sx ) — X Lt sin* (j:+S:>c) sin 8x 0 _ cos X ^ 1 cos X sin X sin x ^sin x — —cosec X cot .X. ^ (cosec 3^)=— cosec x cot x. Alternative Method {noi by ab-initio) Let y=coscc x= — 4 — • sin X dy (sin^) dx ~~ sin* X sin xxO— Ixcos x cos x ~~ sin* a: sin* — —cosec X cot X Hence — {cosec x)=— cosec x cot x. Note. (/) Alternative methods given above are not ab-initio methods. These have been given here for ambitious students. Hi) These results hold good only when x is measured in radians These will not hold good if x is measured in degrees. In such a case x is converted into radians before the result is applied. ' Example. If ^=tan jc®, find dx Sol. Here we have, x degrees^ T^X 180 radians • « 3 >=tan "KX 180 44 A TEXT BOOK OF ELEMENTARY CALCULUS dy n D, dx “180 180 ^ AO = 180 * • Worked out Examples dy Example 1. (i) Find if >>=cos (3;c+5). Sol. Here >’=cos(3x+5) Put z=3,x+5 so that Now y=COS z 4y. dx dz ^ dx ...(0 ...(«) From (i), we get ^—3, and dx from (ii), we get Hence dy sinr dz — — sin (3x+5) dy dy dz ,./*,, c\ ^ (3 x+5). [V z=3Ar+5 dx dz Alternative Method cos (3jr+5) is a function of 3x+5 which is, in turn, a function of x. Hence cos (3 a:+ 5) is a function of a function. [cos (3 jc+ 5)]= — sin (3x+5)x (3jc-h5) ax ax =— 3 sin (3x+5). Example 2. Differentiate tan® x w.r.t. x, Sol. Let tan* x and z=tan x, so that y=z» Now z—tAnx gives and ^=sec* X dx y=^ gives DIFFERENTUTION OF CIRCULAR FUNCTIONS 45 dz =3z*=3 tan® x. dy dy dz Alternative Method 3 »=tan* ;c=(tan xy %=^(\&nxyx~ (tanx) =3 tan® X . sec® x. na/ive^methodT^W^hp"* advised to completely master these Alter- ^ (sin /«x)=m cos mx. (//) ^ (sin*” x)=m sin*”-* x . cos a:. dx sin*"-* nx . cos nx. ample 3. Differentiate tan mx ab-initio. then Sol. Let • • >’=tan mx 3’-t-5>'=tan m (x4 Sat) S>'=tan m (;t+Sjc) — tan mx sin m (.y+^jc) sin mx cosm(x-i-Bx) cos mx _^nm (x+Sat) co s mx-cos m(x-\-Sx) sin mx cos m (;cH-5jif) cos mx sin [m (jc+^x)— wjr l cos /M{jC + 5Aj) ’f~COS'/MX sin m cos mCx-f 5jc) COS mx sin m Sx • • Sat cos OT (a:+SjcJ COS /?j;r sin m m Sx ^osm(x4-Bx)7^n;i^ mx 46 A TEXT BOOK OF ELEMENTARY CALCULUS then or and dy_^ Lt ^ dx Sx-* 0 8x = Lt m sin m m Sx 8.v-».0 cos m (xH-Sx) cos mx m = m sec* mx. cos^ mx Example 4. Differentiate sin x» from first principles. Sol. Let ;»=sin x* >'+S>’=sin (a:+Sx)’^ iy=sm (x+Sx)*— sin x* Now =2 cos 8>* g^=COS sm [ (x+8x)^x* 2 -] ] (x-\-ix)*-x* 2 (x+Sx)*-x‘ Sx Lt cos s^O sm Lt S.r-^0 j- (x+Sx)*- -L. (x+5x)*— x> Also Lt 3.y-»0 (X+S x)»-X^ «x = Lt — ^ 8x->0 6X = Lt 8x-*-0 [ «X Hence -^^=2xcosx*. Example 5. Differentiate sin'^ Zx Vl — x* w.r.t. x. Sol. Let _>-=sin-^ 2x Vl-x* Put x—sin e ...(I) DIFFERENTIATION OF CIRCULAR FUNCTIONS 47 SO that >'=sin‘^ (2 sin 0 VI— sin’* 0) = sin“^ (2 sin 0 cos 0) = sin'* sin 2 6 = 20 . Now (1) gives ^ =cos 0, and (2) gives dy ...( 2 ) ' V 1— cos X l+COS X {K.U. 1976) 3. 4. >.= V COS X 3 tan X— tan* x + COS x \ —sin x) {K.U. Nov. 2975, 1977) 1—3 tan* X 5. y=tan”‘ (sec x+tan x). 6. 7. V 1+COS 3x 1— cos 3x {P.U. 1957) ’=tan~' ( cos X 1+sin X )■ 8 V 1+C08 X COS X 1 — tan* Hint, cos x= 1+tan* X etc. UI. Differentiate the following w.r.t. x : 1 . 1 + x^ 2. cos 4. sin" 5. ,an- ‘ tan" 7. tan-> • 1+X* (K.U. 1975) IV. Differentiate : 1. tan X w.r.t. sin x. 2. sin 3. sin X w.r.t. cos x. 4. sin -1 1-x* 1+x* -d-s) 3. tan“* , i w.r.t. sin 1— X* -1 2x sin -1 1+x* X , 1— X* - w.r.t. COS"* . , '2 Vl+x» 7. w.r.t. tan"* X. 50 A TEXT BOOK OF ELEMENTARY CALCULUS 8 . V. 1 3. sec -1 2 x ^-1 Differentiate : w.r.t. Vl+x*. V sin X. 2. \''tan V X. 4. sin vx. tan V l + .x^. {K,U. 1975} -o VI. If sin j»=x sin (fl4-^), prove that dy _sin® (o-hjO dx sin a VII. If 3'=. /sin jr+ Vsin x+ V sin -Y—to CO C {KXJ, 1976} show that Vlll. Find dv dx , when : cos X 2v-l ■ 1. cos 0, 3'=6 sin 0. 2. x=a cos® 0, y—b sin® 0. 3. x=a sec® 6, y^b tan® 0. 4. .•v=a (0+sin 0), >»=a (1— cos 0). IX. Differentiate ab-initio : 1. cos 3.V. 2. sec 5x. 3. sin Vx. (JC.U. 1976) 4. cos® X. 5. % cos X. (K.U. 1975) 6. tan .X®. 7. X sin X. (K.U, 1975) 8. Vtan X. iK.U, 1975) 3*3. Inverse Circular Functions Def. If sin x=3’, then x is such an ang|e whose sine is equal to V. This can be expressed by saying that sin Again, if sin x=-^ then .visin’* and we know that there . . V3 .rn. Ti 2n In are many angles whose sm is 2 • These are ^ 1 ^ etc. This shows that sin"* x (and similarly any other Inverse Circular function) is many-valued. But we are concerned here only with the smallest value, so far as differentiation is concerned. This smallest value is known as the Principal Value of the Inverse function. 3*4. Differentiation of Inverse Circular Functions (1) Differential Coefficient of sm'^ x. Let y=sin“‘ x x=sin y •••(!) y+5y=sin“^ (x+5jf) x+8x~-sin (y+Sy) •••(2) then or DIFFERENTIATION OF CIRCULAR FUNCTIONS 51 Subtracting (1) from (2), we have Sx=sin 0'+^;') —sin v =2 cos or Sy =cos^ By ) sm sm By By 2 r dx Bx or =cos y dy _ I_ _ / t \ Sin By By 2 1 I Hence ^ (sin** x)= dx ' VI -X- Alternative Method Let ^=sin ^ X x=sin y Differentiating w.r.t. y, we get dx dx cosy vlsin-y V I or d^ ■> 1 I d}\ ^ dx cos y v' l-sin“ v _ 1 X/I_v2 or Hence (sin'^ x)= — - • dx vl-x* (//) Differential Coefficient of cos~^ x. Let ^=cos~* X X — cos V Now _>»+3>;=cos-i fx+^Jx) x+^x=-cos (y-^-By) Subtracting (1) from (2), we get Sx=cos cos y = -2sin(;-+?|-)sin|l 8y ••(I) - (2) or Sin !■>!. 2 52 A TEXT BOOK OF ELEMENTARY CALCULUS r . Sy — sm ( y4-‘^) — ; — dy sy-^o ^y ty^o I \ ^ J ?y_ L 2 =— sm y dy __ 1 1 dx siny VI — cos* y Hence ^ (cos'^ a:) Vl-;c* V 1— cos* y Alternative Method Let y=cos“^ x :c=cosy. Differentiating this w.r.t. y, we get dx ^=^^y dy \ 1 or *7^ = — : — I . dx smy VI— cos* y = — -i — iK V 1-x* , Hence (cos“' x)= . dx V 1— Ar> (/») Differential Coefficient of tan~^ x. Let y=tan“' x A:=tany •••(!) Now y+8y=tan'^ (ac+Sx) or x+8x=tan (y-f"5y) •••(2) Subtracting (1) from (2), we have 8x=tan (y-fSy)— tan y _ sin (y-t-^v ) siny cos fy+Sy) cosy _ sin (y+8i*) cos y— cos (y4-8y) sin y cos (y+Sy) cos y _ sin (y4-8y— y) sin 8y cos (y+Sy) cos y cos (y+8y) cos y sin ^y sf coVo^+S^') cos y ■ ““Sj sin 8y Lt _ 1 dy 8j»-*.o (y+^y) cos y cos* y =sec* y DIFFERENTIATION OF CIRCULAR FUNCTIONS or - 1 - 1_ clx sec^ y 1 +tan' y Hence ^ (tan- » x) = Alternative Method Let >’=tan"^ X •• A — tan y Differentiating this w.r.t. >•, we have dx d}^ 1 1 1_ dx ~sec^ ;»~l + lan“>’~l+A* Hence ^ (tan->x)=^. (iv) Differential Coefficient of cot'^ x. Let ^=cot~^ X *• Jr=cot>’ Now >'+Sy=:=cot“^ (.v4-Sa) ^+Sj:=cot (y+Sy) Subtracting (1) from (2). we get ^A=cot cot y cos cos y sin sin y sin y cos (v+Sy)— cos y sin (y+S>>) sin (>+5^) sin y — sin (y— y+^y) —sin Sy sin sin y sin (y-rSy) sin y —sin 5r sin (>'+8;') sin ;; —sin Sy Lt ~~ Lt dy iy-*o sin (y-\-By) sin y Hence sin^ y dy 1 dx cosec* y d , = — cosec* y 1+col* y 1 1 1-A* •••(0 ...( 2 ) ^(cot-x)=- 54 A TEXT BOOK OF ELEMENTARY CALCULUS Alternative Method Let X x=coty Difierentiating this w.r.t. y, we get dx 3 ^—= —cosec* >» 1 1 dx cosec® ^ H-cot*>» Hence fcot-* (v) Differential Coefficient of sec'^ x. Let ^=sec'^x -x— sec>» Now ;’+5v=scc-* (.v+^.y) or .Y T-S.v=sec (>>+ Sj) Subtracting (1) from (2), we get Sx^sec (>»+S;0 — sec y 1 1 cos 1 1+x® ..( 1 ) ..( 2 ) I 1 cos cos (>>4-Sj>) cos (,v+^v)~cos y“ cos cos y 2 sin ^ ^'+^) sin y cos (>’+Sj’) cos y ( , \ . ^y j8y S'" ( >+r ) S'" T/ 2 dx cos (>'+S>») cos .V sin ( 3 ’+y) si or — T t - ^ - dy ^y_^Q cos cos 3 > ^sin_j' — 1 — — 5CC V tan y cos y ' cos y dy 1 dx ^ sec y tan y 1 sec y V sec®^ — 1 1 "",V V .Y®- 1 Hence " (sec ' -y)= -—===.. dx -Yx/.x*— 1 Alternative Method Let jj^sec"^ .y .Y=sec V 2/2 DIFFERENTIATION OF CIRCULAR FUNCTIONS Differentiating this w.r.t. we get dx ^ = sec>'tan y. dy 1 1 Hence dx sec y tan y sec y V sec" y — 1 I dx X ^ 2 _ 1 (sec"' A') 1 XV A-2- 1 (vi) Differential coffecient of cosec^ x. Let y=cosec^A .v=cosec y Now y+^y=cosec“^ x+5x=cosec (y-f 5y) Subtracting (I) from (2). we get ^A:=cosec (y-r-Sy)— cosec y 1 I sin (y+rfy) sin y sin y— sin (y+^y) sin (y -j-Sy) sin y 2cos(j.+^^-)sin (-1) sin (y+'5y) sin y -2 cos >■ f sin ^ sin (y+'^'y) sin y dy sin (y -t-Sy) sin y — = Lt dy jiy^Q ^y sin (y-f-^y) sin y cos V 1 sin y ■ sin y = — cot y cosec y 56 A TE?Cr BOOK OP ELEMENTARY CALCULUS or dx cosec y cot y 1 Hence cosec y v cosec* y— 1 1 xVjc*— l’ |^(cosec-»A;)=- ^^j_l Alternative Method Let >>=cosec“^ X cosec y Differentiating this w.r.t. y^ we get dx =— cosec y cot y. • • dy dy^ dx 1 cosec y cot y 1 Hence Example 1. If cosec y v/ cosec* y— 1 1 xy/x^~l 4 - (cosec“^ jc)= ^ dx Xy/j^—\ Solved Examples ^=cos^* (sin x), find ^ • Sol. Let z=si so that sin X y=cos~^ z dz Now from (i) -y- =cos x dx and from (//) dz 1 Vl~z* Vl— sin*x dx^dz^dx = — — ^ — xcosx= — 1 . cosx Example 2. Dificrentiate sin"* b+a cos X 0.^1 T ^ *1 a+b cos X Sol. Let >=sin~* -H o+acosx -.71 ^ a 1 cos X DIFFERENTIATION OF CIRCULAR FUNCTIONS 57 and _ a+b cos X b-ra cos a: z= so that y Now form (i) dz dx sin”^ z. ■•■(/) and from (n) az ~b sin X {b-^a cos a )+<2 sin x{a-\-b cos x) (6 + fl cos a:)" ~~ — jb^—g-) sin x (6+a cos x)^ 1 1 V <3 + 6 COS X V b-\-a COS X b-\-a cos X 1 ^{b- a cos x)*(— 0+6 cos a)® 6 +o cos X V ( 6 ^— 0 ")— ( 6 ^— 0 ^) COS" X 6+<3 cos X dy_ dx o* sin X dy dz 'dz ^dx b-ra cos X \ b- — Q^ sin X I o o y b-—a- X (6^—0^) sin X ( 6+0 COS x)* 6 : o cos X Example 3 . If sin"‘ X V 1 — X .2 , show that Sol y= sin"* X V 1-x* Differentiating both sides w.r.t. x, we get 2x>'i(I-x3)-2x;»2=2 or ;'i(l-x*)-x;^=l. dv sin"* X V l-.r* = 2y (Please note this step) (Please note this step) Hence (I -X*) ^=ix-xy. 58 A TEXT BOOK OF ELEMENTARY CALCULUS Exercise VIII I. Differentiate the following w.r.t. x : 1. sin-1 3x. 2. cos-1 (3x+2). 3 J, , 4. sec-i x^. 5. tan-i (tan x). 6. tan (tan-i^). 7. cosec-i v'x. 8. tan'i x+tan-i — • X -1 cos X cos 9. tan x.tan-i x. 10. II. Differentiate : I. sin-1 _L ^ j. t tan-i (2.Y+3). ^ V a+h cos x‘ March 197S) 2. sin* X , 1 — Y* — == w.r.t. cos-1 V 1 -tx^ 1 +A* 3. tan-i w.r.t. sin-i ^ III. 1. Differentiate both sides to show that (0 3 sin-1 ;v=sin'i (3.x— 4 a-®). 2 .x (K.U. 1976) Hi) 2 tan-i A=tan i 1-.X® 2. If ,v=sin-» .x-^sin-i Vl— prove that £ = 0 . [Hint. 3. If j>=sin-i .x, prove that d.\ “ dx 4. If >»=cos (m sin'i a), so that n -*■> S-' % IV. Differentiate W.r.t, .Y : sin*^ V 1 — =cos"^ X ] sin [ 2 tan- [Hint. Put A=cos 6.] tan 2 . sin- >'‘ + ^+^^--^ . [Hint. Put A=cos 0.] '-[VS tan fj 4. -X COS'l .X l-.Y® (P.l/. 1942) V. Show by differentiation that ,-i 2 a 2a 1 yS -=-^ — l-sin“i is constant. {K.U. Nov. 1975) [Hint. Put .x=tan 0]. 4 Differential ion of Logarithmic^ Exponential and Hyperbolic Functions ‘^’1. Before attempting to difl'ereniiate logarithmic, exponen- tial and hyperbolic functions, we shall find the limit of ^ 1 j as n tends to infinity. With the help of this limit, we shall next proceed to find the limits of {1 and ^ ^ as .v tends to zero. These X two limits will enable us to differentiate the functions referred to above. 4’2. To show that Lt f 1-r — ) —e 2+t-!( *-v)+3-t( '-tX' 2 ! n(n~])(n- 3 ! + •• As n oo n 2 n Lt , all -0. l+H-~-r to «> ? * n-*-0 * fl*— l=z, so that z-^0 as x-^O o*-l=2 (2*=1+Z log o*=log (1+z) X log a=log (l+z) logfl+z ) log a ^ r^O ^og (1+z) (14-r) log a iK.U. Nov. 1970 z log a _ log a log a Lt i log (H-z)~ Lt log(l+z)i'* r-+0 r-^0 log a log a 7.=^=IOgO. logLtd+zP'* log e r-^0 Cor. Put Lt ^^=1. iK.ir. m 7 > a=e, then log e=l. Or Lt n-»0 ...(/> Put e*-l=z e'=l+z z->0 as .x-^-O log c*=log (1+z) x=log (1+z) 61 DIFFERENTIATION OF LOGARITHMIC FUNCTIONS or (0 becomes : Lt *= Lt 1 £->•0 (l+z) I log (l+z) = Lt = !og(l+z)~ z Note. In the foregoing article, we have applied an important iormula of Trigonometry pertaining to logarithm. The formula is : «oga loga m. The student is advised to note this carefully. He IS also advised to note that : (1) log,, /Mn = loga w + log, n. (2) loga /w/n=log« m— log. w. (3) log. 7M=log^ mxiog. h= 1°^ ■ (4) (5) \og„a=l. Napier Logarithms In Trigonometry, we take 10 as the base of logarithms. In ■Lalculus, however, we take e as the base, so that when no ba« is mentioned we take it to be e. Thus log X always means Iog« x. fhms ^ their base are known as Napiers Logari- ^ 4. Find the Differential coefficient of log. x. Let y=log. AT 3’+5y=log. U+Sx) 5y=Iogo U+5 a)— log. .V AT+Sx then or =log<. ( V log. m— logo n= = log« ( 1 + 5x ) -io.. -= ) ^=S-Iog.( l + -j or 62 A TEXT BOOK OF ELEMENTARY CALCULUS 1 X X Sx lOga ( 14- = ^10ga( 1 + ^) Sx X ax ) : ' .‘Ts- jyi ,* fPlease note these two steps! dy_ dx Lt 8jr-»-0 X 8x--*0 Lt lo&.( 1+^) c-^0 \ X / X 8x 1 = — log., Lt X ® 8x^0 (- 1 ) X ax Now Lt ( 1+^^ Sx-^O V ^ X a.v ) — 1 f / Lt (1+x) ^ —e I L x->o J Hence ^ (logr. logae. Cor. If a=e, then" (logx)=-^ log, e _ j_ ~ X Note. The Cor. under article 4'4^ gives us a very important working rule differentiating Logarithmic Functions. The rule can be laid down as under * Differential coefficient of the log (of some function of x) % i- X Differential coefficient (of that fwtction of x) that function of x w.r.t. X. This can be illustrated by means of the following solved examples. Example 1. Differentiate log 3x with respect to x, Sol. Let y=log 3x =-^ xDiff. coefficient of (3x) dx 3x =A_i. 3x “■ X Let so that differentiation of logarithmic functions Alternative Method : ;’=log3jc and z=3:c 3 -log z. Now v=iogz gives az z 3.r and = 3.v dz _ gives ^ = 3. dx ~dz ^dx-3x ;c ' Example 2. Differentiate log (sin at) w.r.t. x, Sol. Let 3 '=log (sin xf dy 1 . d , . , COS X sin X = COt X. Example 3, If y~ log tan -r- , show that I* dy dx = cosec X. Sol. Here y= log Ian A' dx tan 1 d X ) 1 2 — sec® -:r- 2 2 X tan y 1 --- 2 X 1 cos X cos sin 1 1 ^ . X X 2 sm -y cos y Sin X cosec A. ^'5. Differential coefficients of a"**. Let then or } Zy = a _ IaH-Sx) _ ^n7 + (.x + ix) gtfix — 1 ) 63 64 A TEXT BOOK OF ELEMENTARY CALCULUS or 8jL Sx ^ Sx -“'“x m6x [Please note this step] 4 - Lt ^=ma’”'x Lt dx 8,v->0 8x-M) mSx But -mSx 1 / Lt — =loEQ (v 9x-^0 Lt '^^=®loga) x-t-O * / • • • log a. dx Hence d iQg ^ dx • Or Differential coefficient of (a"**) X Differential coefficient of {mx) X log a. Example. Differentiate a®' w.r.t. x. Sol. Let y—e^. (3x)xloga dx ax =3a®* log a. Cor. Differential coefficient of w.r.t. x=me'"*. In this case a=c ^ (e”**)=/MC’"* log e=me’"'' (’•' loge=l) Hence ^ log a. 4*6. Alternative methods for differentiating a"® and e"^ w.r.t. X. (i) Let y=ar^ log y=log fl”*® =mx log a. Differentiating both sides wj.t. x, we get 1 dy , — =m log a, y dx -^-=my log a=-ma”'^ log a. dx ' 65 DIFFERENTIATION OF LOGARITHMIC FUNCTIONS <») Let y=e' ]ogy=loge' =mx log e=mx Differentiating both sides w.r.t. x, we get y Hence dx dy dx m =my=me mx ( .* loge=l) Note 1. The differential co-efficient of log y w.r.t. x 1 j 1 y 2. The methods explained under Article 4*6 are not ab-initio methods. Estaxnple 1. Sol. Let and so that Solved Examples Differentiate log (log x) w.r.t. x. y=\og (log x) 2—\og X >'-l0g Z Now (1) gives dz l_ dx ~ X and (2) gives l_ dx z ~ log X Hence d^ _d^ t/z _ 1 J 1 dx dz dx ^ log a: ^ x ~x log x Example 2. Differentiate ^ w.r.t. x. Sol. Let AT and let 2= tan X so that l! Now from (IJ (!z j — =sec* X dx and from (2) dz « « • dx dz dx - scc^ X 66 A TEXT BOOK OF ELEMENTARY CALCULUS .tan A' Alternative Method : Let y=e Taking log both the sides, we have log;j=e*®" ^=tan x log e = tan -Y. Differentiating this w.r.t. .v, we get 1 dy „ — 7^ =sec'* -v V dx Hence dy T- = v dx sec^ .Y=sec- .v e tan -Y. uV Example 3. If v- log (.v+ V 1 find Sol. Let so that Now =^X-Vl-r-r- j-log z ^ _-liL ^ ‘iL _ J_ V —( dx dz ^ dx z dx V .V f- V H-.v* ) ..(2) 1 .Y-+- V A-S+l^l^^ V.X-S-f l) X-, 1 -’.X'^ 1 ^ V U-;- V X^'4-1) ^ Exercise IX 1. Differentiating tlie following w.r.t. .v ; '' H-x*+x : ^2 \ 1. log '^/ 3. log tan ( ^ . 1 - cos .V *• '°E 7. log tan’^ -Y. )• 2. log 4 V l + A^-.V ■ log (sec AT -f- tan .y). 9. f”' 6. log (.V tan x). 8. log (log a). . X 10. log than + 2) {K.U. 1957, 77) 11. Differentiate the following w.r.t. a : 1. at'. 2. t>"*+e2'. 4. cosle^ + c-'). 5. sine* 3. 6 . X tan X 7. 8. ^ 10. log (xM e*). X tan"^ X tan C2.Y+ 3) a DIFFERENTIATION OF HYPERBOLIC FUNCTIONS III. I. If log = show that ^ . 2 :y -— 1 dx X l—2y^' 2. If >’=Iogj; X, show that ^ 1 dx ~~ x(l-Mog y) 67 (J. & K. 19i>0) 3. If Vlog x+viog ^-rVlog :c+T.to oo show that ^ = -p: -- dx xi2y—l) 4. If 10, show that dy log 10 dx ~ X (log x)* 5. If >'=log show that ^ lo g X dx ~ (1 -flog IV. Differentiate the following w.r.t. x : I 1 . tan*' (log x). 3. sin (iog 2. sin (log tan x) (K.U, 1954) {K.U. Now 1975) 4. log x*-f 1 Vi^ I *> x-\- x- x-\-x 2 5. a log a — b tan x 7. log [log (log x)]. 4*7. Hyperbolic Functions is defined as the Hyperbolic sine of x and Definition. e^—e‘* ^ - Xi- XJJ UflQ is uS* "*' hyperbolic functions are defined 1. Hyperbolic cosine of written as cosh x. 2 e—e~* 2. Hyperbolic tangent of x= ^ written as tank x. 3- Hyperbolic cotangent of written as co//i x 68 A TEXT BOOK OF ELEMENTARY CALCULUS 4. Hyperbolic secant of written as seek x. 2 5. Hyperbolic cosecant of « written as cojecA x. From the above definitions, it is easy to see that : sinh X 1 . tanh x= cosh X . cosh X 1 3. sech x= 4. cosech x~-- cosh X 1 sinh X 4-8. Some important relations between Hyperbolic Fano tions 1. cosh^ X — sinh- x= 1. 2 . 1— tanh“ x=scch“ x. 3. coth* X— 1 =coscch- x. Proof. 1. cosh® X— sinh* x=l L.H.S.=^(cosh X i sinh xKcosh x- sinh x) e*-e 2 A 2 2 Relation (2) can be obtained by dividing relation (1) by cosh* X. and relation (3) can be obtained by dividing (1) by sinh x. 4'9. Differentiation of Hyperbolic Functions (1) Let vs=sinh X— — ^ • • • dx 2 Hence ~ (sinh x)=cosh x ux (2) Let y=cosh X— — 2 + sinhx dx 2 Hence 4— (cosh x)=sinh x. d.x DIFFERENTIATION OF HYPERBOLIC FUNCTIONS 69 (3) Let (4) Let >'=tanh x = sinh X cosh X dy cosh^ jc — sinh^ x dx (cosh x)- 1 cosh^ X =sech^ X, >’=coth x= cosh X sinh X d^ sinh X . sinh . v— co sh x . cosh x dx “ (sinh x)* c osh^ X— sinh^ .v 1 sinh® X sinh® x = — cosech® X. (5) Let >'=sech 1 cosh X dy Ox cosh x— 1 xsinh x dx ~ (cosh X)® sinh X 1 ” cosh X ■ cosh X = — sech X tanh x. (6) Let >'=cosech x=-r . sinh X . ^ 0 Xsinh X — 1 Xcosh x dx ~ (sinh x)* _ cosh X I sinh X ■ sinh x = — cosech X coth x. Note. The student is advised to mark the difference between the ^Ivatives of circular functions and those of the hyperbolic functions More or less, they follow the same order, but the last three derivatives in the case of hyperbolic functions are negative, whereas in the case of circular functions, these are alternately positive and negative. '' 4* 10. Differentiation of Inverse Hyperbolic Functions (I) Let y=sinh-^x or x = sinhy . dx di y dx “"cosh y 1 1_ V 14- sinh* y V 1 +x^‘ Hence 70 A TEXT BOOK OF ELEMENTARY CALCULUS or or or or (2) Let cosh“^ x x=cosh y "sinh y Hence dy dy_ dx 1 sinh y 1 1 x/cosh*^— 1 Va'*— 1 (3) Let y=tanh *^ x y jc— tanh y Hence d^ dy dx , <, 7 — =secn- y 1 sech* y 1 1 1 — lanh^y 1— (4) Let y=coth-‘ x .x=coth y dx dy 41 dx = — cosech® y Hence -r^ = - 1 cosech* y t 1 coth* y— 1 (5) Let y=sech"‘ x sech y Hence -r- = — y X dx dy 4t dx = — sech y tanh y 1 1 sech y tanh y I sech yVl— sech*y (6) Let or ~ xvl— .X* y=cosech'‘ x x=cosech y dy 4y dx T- = — cosech y coth y I coscch y coth y Hence DIFFERENTIATION OF HYPERBOLIC FUNCTIONS 71 1 cosech yy/ cosech’ >’+ 1 _ 1 x\/x^-^\ Note. We have knowingly avoided writing ± signs outside the radicals, and the students need not go into such niceties at this stage. 4'11. Logarithmic DilFerentiatiozi In order to differentiate a function of the form m‘, where u and V are both variables, it is necessary to take its logarithm and then differentiate. This process is known as Logarithmic Differentiation, and IS also useful when the function to be differentiated is the pro- duct of a number of factors. The following solved examples will illustrate the method : or Example 1. Dift'erentiate (tan w.r.t. x. Sol. Let >>=(tan .t)'*" log .v=Iog (tan ' = sin X log tan x. Differentiating w.r.t. x, we get Hence Example 2. Sol, Let dy • dx =cos A' log tan x-hscc x =cos X log tan x+sin x . sec^ X tan X (cos X log tan x+sec x) - (tan x^'" ^ (cos x log tan x-hsec x). Differentiate x'**" ' -f(sin x)*^”^ >'=x'^n ^jcos .V and ^ v=(sin x)‘'^=* ^ 80 that y===:u -^V ^ __du dv fix dx dx From (1) taking logarithm, we obtain, log tt=tan X log X 72 A TEXT BOOK OF ELEMENTARY CALCULUS i.e. or I du ^ . . 1 X log x+tan x . ~ du tao X dx = X (sec>xlog;c+i^) From (2) taking logarithm, we obtain, log v=cos X log sin x 1 dv • 1 • I cos X ~ = — sin X . log sin ;c+cos x . sm X = — sin jc log sin x cos* X sin X dv . .cos X Fx f— sin X log sin ^ ) \ ® sin X / Hence. dy _d^ , rfv tan X dx ~dx dx ~ ^ sin X tan X ^sec* X log xH-^^-^ ^ + (sin x) COS X (-si sin X log sin x+ X cos* x \ sin X / Example 3. Differentiate Sol. Let y= x*'*(l-2x)*^* (2-3x)*«{3-4x)*^® x*/*(l-2x)*'* (2-3x)*'*(3-4x)*'^ Taking logarithms, we obtain log >'=i log x^§ log (1— 2x)— J log (2-3x) log (3-4x). Dififerentiating, we get dy _\ 1,2 -2 3 —3 4 -4 y ‘ dx 2 • X 3 ■ l-2x 4 * 2— 3x' 1 4.9 16 ^ - dx 5 • 3-4x 2x 3(l-2x)'^4(2-3x)^5(3-4x) X 3(1 -2x) ‘ 4(2-3x) -2x) 7k + 16 ~ 5(3-4.v;. (2-3a-)=‘/^(3 >x)- ^ r 1 -4x)^'" L2-’ + 2x 3(1— 2x) ‘ 4(2“3x) + 5(3 16 ' -4xL u Example 4. lfv=—, where « and rare both functions of X, show by logarithmic differentiation that dx du dv V ;y U y- dx dx ..2 DIFFERENTIATION OF HYPERBOLIC FUNCTIONS 73- Sol. Taking logarithms of both sides, we get log >=log ^ =log w— log V. Differentiating w.r.t. x, we obtain 1 ^ du 1 dv y ' dx~ u ' dx V ' dx Exercise X (Section A) Differentiate w.r.t. x : 1. sin X log X . e‘ . y/x. 2. v' X . Vsin X . Vlog X. 3. 4. (tan 5. (sin"^ x)*. 6. ysmx 7. (sin xi'-s ■' . 8. cos (x^). {K.U. 1977) 9. 10. X --' . 11. X i x>'^ 12. (sin xf^^ ^+(cos x)-"^ \ 13. (sec xY^^ * + (coscc xT'^^ ^ . K (tan 2* cot jf ^ cos'^ X VJC ■ ■ V l-;e 17. If y=iA', where u and v are functions of x, show by logarithmic differentiation that ^ + ^ dx dx dx 74 A TEXT BOOK OF ELEMENTARY CALCULUS 18. If to oo 19 20 21 22 23 prove that dy y ^dx j-log X flv(cos 6) to oo If j^>=(cos ' , prove that dy >’* tan 0 y log cos H — 1 If x-^ = ’=j: log r, prove that = py-.p~v dy 1 If A-= . show that = ir e' + e*' If X— — ^ ■i Tc - * 1 . * dy — , prove that V 1-rJt oy , » Section (B) Differentiate the following ; , . , , 1 • ua COsIx + COShX 1. smh x + i smh-* .Y, — ; r — • sinlf x + sinh x sinh x 2. log cosh X, log tanh x, a 3. sinh'^ Vx^— 1, sinh-^ (tan x). 4. see"* (cosh x), tan"^ (sinh x), cos"^ (sech x). 5. 2 tanh"* (tan ix), cosh'* (sec x). (P.C/.) iP.U.) {PV.) iP.U.) 5 Successive Differentiation 5*1. Introduction. The derivative of a variable func- tion /(r) is itself a function of .v- We suppose that it also possesses a derivative which wc denote by /T.r) and call the second derivative of fix). The third derivative of /(x) which is the derivative fix) is denoted by/'"(x) and so on. Thus the successive derivatives of /{x) arc represented by the symbols /'(x),/'’(x)./"'(x), /"(x) where each term is the derivative of the preceding one. Alternatively, if y=fix), then ^ denotes the nth derivative of y. Sometimes yi, y-., y., y„ are also used to denote the successive derivatives of y. Solved Examples Example 1. Find the third derivative of 5x^-l-6x‘-t-3x“+l. Sol. Let y=5x^-!-6xM'3x2+I. Denoting the first, the second and the third derivatives of y by yif ya yz respectively, we have - 25x‘-|-24x" + 6x >-2=100x^4-72x2 L6 and y3=300x*4-144x. Example 2. lf>; = sin (sin x), prove that d^y dv j-j4-tan X — 4-y cos^ x=0. {K.U. 1977) Sol. Let >’=sin (sin x) 1 # ^ =cos (sin x) . cos x d^y ^ = — sin (sin x) . cos* X— cos (sin x) . sin x 76 A TEXT BOBK OF ELEMENTARY CALCULUS Making substitutions in (1), we get —sin (sin x) . cos* x— cos (sin x) sin x+tan x X cos (sin x) cos x+sin (sin x) cos* x»(k Example 3. If x=/(/), >’—9(0. prove that dx d*y dy d^x d^v dt ' dt^ ~dt ' dt* dx^ (^fJ {K.V. i975)- 4y, Sol. dx ^ dt Differentiating both sides w.r.t. x, we get dx^~dx\dx J^dt \dx dt dx (Please note this step) 4L dx dt tPx dt^ if) 1 Xj- dx dt d-y dt- 4l. dt 4y dt d‘x dt- (r) Note. It has been observed that students generally commit mistakes while solving questions of the type of solved example 3. They commit mistakes particularly in respect of the deno- minator of the above result. They are, therefore, advised to take this example as an article and understand it as thoroughlyjSas possible. Example 4. If .Y = rt 01 sin 6). v = a (I — COS 8) find Sol. Hence ^ (I — cos B). ^ — d'f «'♦ =a sin 8 dy (h d<> a sin H H f! 2a sin -j- cos ^ dx dx — COSO) 2a sin* It 2 =cot 2 SUCCESSIVE DIFFERENTIATION 77 -Again, dx\dx ) dx d ( fi \ d^ _i_ 2 J_ Aa cosec 9 , 9 cosec* -j- 1 <7(1 —COS 0) 'Example 5. If 3 ;=sin {m sin'^ x), prove that Sol >'=sm (m sin*' x) ^ =cos (m sin"' x) . dx Vl-x" Squaring and multiplying cross-wise, we get (1— x') ^=m- cos^ (m sin"* x) — [1 — sin" {m sin“' x)) -m2 (I-j2) Differentiating both sides w.r.t. x, we get or cancelling out 2 ^ , we get It should be noted that the differential Exercise XI Find the 1. Fourth derivative of x*. 2. Third differential co-efficient of ~ • X 3. Fifth derivative of — - — ox -1-6 4 . Second derivative of log (3— x). 5. Third derivative of tan x-f cot x. • -( 1 ) [from (1)] coeflBcient 78 A TEXT BOOK OF ELEMENTARY CALCULUS Second differential coefficient of If show that = log X . ^ d‘^y 2 log x— 3 If y= show that If y=^log (sin x), show that >> 3 = — ^ ® sin^ X If y=A'+tan x, prove that cos- .V r*' — 2 v+ 2 a= 0 . dx^ If .v = A cos nx - B sin nx, then V , / vM-; .V--1 It r-^tan-^[ j tan 1 2.V l-x- show that vuiaw )■,- [Hint. Pat x = tan 6, etc.] 13. If x=(a ~bt) show that 14. If .t=2 cos / — cos 2/, V— 2 sin r-sin 2/, find d'Y , . ^ dx^ 2 d'^ ^ 15. If.x = a cos fJ, y — 6sin0, find 16 . If .v = sin /, y^sin pt, prove that 17. (0 If y^= (sin - ^ .v)^ prove that ' ( 1 — A=)y2 — xyi'=2. Xii) If y=sin (sin x). prove that .V’-rVi tan x+y cos^ .v — 0. 18. If ky=s\n Lx r y) where k is a constant, prove that y--^— rd-f-yO^ 19. lfx--cos6, y — sin^ 0 show that y y“-3 sin^ e (5 cos* 0 - 1). {KM. Nov. 1975) SUCCESSIVE DIFFERENTIATION 79 5*2. Standard Results Below we establish a number of standard results which often enable us to find the «th derivatives of various functions. The student is advised to have a clear graph of these results in his own interests. j’= (aA‘+6)''', to show that tn ^ Here y=.{ax-\-by'' ;'i— y2=m(ni — 1 )a\ax-r ;’3=m(n7 — ])(m — 2) a^(ax+b)”'~'^. So that in general y. mint— 1 )(m — 2) (m—n 4- 1 )a^{ax f If m is +ve integer, this result can be written as Cor. 1 . xhis /?;th derivative of {ax-\-b)'” is m ! o'" (which i« constant). i- Cor. 2, If y—x", then y.^ n ! y'=- — —7, to show that ax-rb ^, - (-1)" \ji_ 0- (ax-rby*^ Here y={ax+h)-^ y^^—(^ax-\-b)-^ >'2=2a-(ax-l- b)'^ = (-1)2 I aHax+b)-i^*^i >'a= —6aHax-yb)~‘^ = (- 1 )=* Hence, in general Cor. If 3. If X (-1)" I n .n •f^L y=log (ax-\-b), to show that {ax-ybr a 80 A TEXT BOOK OP ELEMENTARY CALCULUS then then Here >'=log iax-\-b) % • y^=-aHax-\-b)-^ =(-!)> |J_a2(ajc+6)-® >'3=2a®(ax-ri)-* = (-!)* Hence _V„={ — 1)"-* |n— 1 a”iax-\-b) (-1)"-^ In-l or iax-^br Cor. If ^=log JC, (-ir-» i«-i 4. If to show that (log u)^ Here % ♦ 4 >»i=wa"* log a (log aY >- 3 =m=*a"’"^ (log fl)^ Hence Cor. If (log a)\ v=e'"% This result can also be established by putting a—e in 4 above. 5. If sin (fl.x+A), to show that sin >'=sin {ax-{-b) yi = fl cos (ux-l-6) =a sin ^ ax+b Va=a* cos {ax-\-b ( rr *2 ) 1^ sin ( +® )=cos 0 J Here SUCCESSIVE differentiation 81 = a® sin (^ax+b-\-~-\-~ ^ = a^s\n (ax+b-\-2 yj =a^cos ^ =a^ sin ^ =a^ sin ( axH-6+2 f) ax + b+^-\ then Hence _Kn=a" sin ( ^x+A + y ) Similarly, if >'=cos (ax >rb)y yu=a'‘ cos ax-\-b-\-~y This is left as an exercise for the student. Cor. Ify=sin.T, then .v.=sin 6. If Here Put so that y^e'*^ sin (/>.x+c), to show that y,. = (a^+b-‘r“ e" sin ( bx+c+n tan-‘ >>- e"* sin (6x+c) yi=ae’‘^ sin (bx-\-c)-{-be^^ cos (6.x+c). a—r cos 0 and b=r sin 6 r=(a®+fc*)i/2 and 0=tan-> — yi=re^* [sin (bx-\-c) cos e+cos (6x+c) sin 0] =re^’ sin (6x-hc+0). Similarly, y^^ r^ sin (6x+c+20) yt=r^ sin (6x+c+30) Hence .>'«=#■" c“* sill (6x+cH-nO) =((i*+6*)"/s sin ^ 6x+c+/i tan-i yj 82 A TEXT BOOK OF ELEMENTARY CALCULUS Similarly, if y=c« cos (6.v+c), ‘hen = e-' cos ( bx+c+n tan"' )■ This is left as an exercise for the student. 5'3. Determination of nth Derivatives of Algebraic Rational Functions. Use of Partial Fractions In order to find the ;/th derivative of an algebraic rational f unction, we have to decompose it into partial fractions. The method is explained below by mcai-.s of a solved example : lx — 1 Example. Find the nth derivative of Sol. Let so that Putting 2.V-I (-V-2)(A'-f 1) B (.t-2X.r-{-I)“x-2 ' .x+l 2x-l = A(.v + l) + B(x-2) .v==2, we get A = 1 Similarly, putting x= — 1, we get B= 1 Hence Now if .v=-k+ ' X—1 Arl 1 , (—1)” I " a' ...( 1 ) Hence (Ogives + Solved Examples Example 1. Find the nxh differential co-efficient of I 3.r+2 Sol. Comparing it with 1 , we find that a=3, b=2. ax-\-b the nth difierential co-efficient (-1)" l.l. a" (-1V !.^L 3 Alternative Method (3.x + 2)"-^i Let yj=_3(3A-+2)-2=(I-I)* _L 3'(3.v-i-2r<'*'» v,---I8(3.y-|-2)*=*=(-1)2 'J 3M3.y ; V3--162(3a- ;-2)-‘-(-lX 12. 3*(3.Y4-2)-f'+») SUCCESSIVE DIFFERENTIA! ION 83 Hence >-« = (-!)" \JL 3"(3x+2)-('+"> (-1 )” \JL 3” (3;c+2)’‘+i Note. Both the methods are permissible, but the latter should be preferred to the former. 1 Example 2. Find the «th derivative of ^ \ZX “T* >'i=-4(2a:+ 3)-»=(-I)* |_2. . 2^ . (2x-|-3)'ti+2) >-2=24{2x+3)-‘=(-ir- 1± . 2" . (2x+3)-(=^^> Hence 1)” |^L±J - 2" . (2;t+3)-‘"^’‘> _ f— 11’* l«-H 2 '‘ ~ (2x+3)"-^2 Example 3. Find the wth differential co-efficient of 1 fl^'+e-’ + Iog (ax+b) + ax-^b Sol. Let >’=fl'‘"'4-e’^ + log (ax+b) H ax-\-b Now y^=ba*‘^ log a+e"'4 ~.-\-i — a)iax'^b)'* ox-ro = ba'" log a+ae^'-f ( 2 (axH- 6 )”^ + (-l)‘ \± (ax+b)-^^*^> a'^dog a)“+aV+(-fl")(ax+6)-‘* + 2a\ax+b)-* = 6 V^(log a)^-{-a'^ e“'+( — O’* |J_ a‘* (ax— + (-1)2 \2_ a2 (o;f+^)-a+*) y 2 =b^ a'^^dog a)"+aV'+2a=*(ax+6)-» ( — 6a^)(ax+6)"* = (log a)®+aV' .(-DMA (-DMA (ax-i-bf iax-rb)^*^ Hence, a''* (log a)" + a" e"' (-ir-* g" , (-ir 1 1. o’* (ax-|-6r + (ax+Z>)'‘+i 84 A TEXT BOOK OF ELEMENTARY CALCULUS Example 4. Find the wth diflerential co-efficient of sin X cos 2x. (P.U. 1957) Sol, Let j=sin X cos 2.V = i (2 cos 2x sin x) = i(sin 3x— sin x). Now each is of the form sin ax whose wth diff. co-efficient is - / , ari \ fl" sm I flxH — !• Putting a— 5, a=l respectively, we get sin ^3x+-^^— sin Example 5. If _)-=cos“ x, find .v«. {K.U. 2976) Sol. Now y— cos® x=i (4 cos® x) =J(3 cos x-i-cos 3x) [ V cos 2A= 4 cos® A— 3 cos A] Now each is of the form cos ox whose wth diff. co-efficient is a" cos Putting a= I and o=3 respectively, we have .v„ = 'j^3 cos ^x-f-^^ + 3" cos (3x-|- Example 6. Find }'» if sin 3x sin 4x. Sol. Let y=c** sin 3x sin 4x ==ie=“'(2 sin 4x sin 3x) = Je®’’(cos X— cos 7x) — lie*' cos x—c** cos 7x]. Now the term in each part is of the standard form “c"' cos bx". yn=i(2*-r 1®)”'* e*' cos (x4-« tan’* i) ^{2* + 7*y“* e*' cos (7x+/i tan"* i). Exercise XII Find the nth derivatives of : 1 . 3. 5. 7. 1 + x 1 a—bx' 1 (2x+3)® * a sin® x-\-b cos* x. 2 . 4. 6 . 8 . 1 a—x x-f2 3x+7’ sin X sin 2x. (K.U. 1976) SUCCESSIVE DIFFERENTIATION 85 9. cos ;)c cos 2x cos 3x. 10. sin^ a*. 10 (a). show that (x+4)"+“ J‘ ^ L(x+2r+i 12. e* cos X cos 2x. 14. X cos a 17. 19. 16. e* sin2 X. 18. sin V^x. 20. sin 4x. cos (x sin a). iP.U. 1958, 55) 11. cos* X. 13. ^ (cos x4-sin x). 15. sin^ X. e* cos® X. e^* sin 3x cos 2x. 5-4. Leibnitzi’s Theorem for the nth differential coefficient of the product of two functions of x. Statement. If where u and v are the functions of x having derivatives of any desired order, then ^ ^ ^ >*’‘="^0 «„v+"Ci u„-i Vi4-"C2 U„-2 Vg-f , . , "CrM,t-rV, + ”CnUVn where u. and v, denote the nth differential coefficient of u andvwrt x respectively. ^ We shall prove this important theorem by the method known as Mathematical Induction. Let us suppose that the theorem is true for any particular value of n. This means that this equation is true for this parti- cular value of n. Differentiating this equation once again w.r.t. x w© set * y-.n="Co(««nV+w«vO + "Ci(«nVi+w,.-iV2) + "C3(M,..v..H- Wn- 2 Va) + +'‘C,(««-f+iVr + w„_^+,)4-..r... +"C„(«iV„-fwv„fi) + + ( Cr)w,.-H-iV,-l- 4 -"Ci.«V„4j ...(1) Now "Cr.l4-'’Cr=''+*Cr Putting r=l, 2, weget "Co+"Ci="+*Ci and ”Ci + '*C 2 ="+*C 2 and Also "Co=I = "+*Co "C.-l="+iC„+, Hence (1) becomes y,.+i="+>Co«,,+iV+"+'C,M„v,4-"+'C2U„_iVa-f + ^*^C,Un-,-^x^'f-\r 4 - This shows that the theorem is true for m 4-1 w^ii u f«* 0+ iTs welt °f "• "rue 86 A TEXT BOOK OF ELEMEFJTARY CALCULUS Again, by actual differentiation, if y=uv then + and y2 = W2>' + WiVi + »iVi+w>‘2 = W2V-t-2w,Vi + «Vj = "C0UV.2+ *CiWiVi -h 'C2MV2 This again shows that the theorem is true for n = 2. Hence it is tn e for i.c. for n = 3 and, therefore, for i e. for n=4 and so on. Hence we conclude that the theorem is true for aXiposiUve I'/jft’gru/ values of rt. Solved Esampl^s and Example 1 Find the «th derivative of -xV. Sol. Let v=xV', and «=e* V - .V*, so that =''0/'a-+"CiW«.iVi+''C2U.,-2Va+ gives y„="Co utx®==v. The student is, therefore, advised to take that function L u which can easily give him the nth derivative. He is at liberty W make a free selecuon in this behalf if both u and v can g.ve nth derivatives with equal case. Example 2. If prove that (l-x-)A+2-(2/i + l)A->Wt-(n=*+o=)y«=0-^ or or Sol. Wehave ■*.—== V 1 ” >V(l-x^) = ay Differentiating again, we get v.t I - X-) - xvi — 0 ay V SUCCESSIVE DIFFERENTIATION Theor^m^'w^get'"® '' times with the help of Leibnitz's >\.42( 1 — .V-) +''Ci,r„ f ,( — 2x) 4- "Q rn( — 2) xyuM-*'Ci.i.y—a^y^=0 which on simplification, gives ( 1 — x 2 )>>„+ j — (2/? + 1 )x v„4i — (/r 4- 0 . Example 3. Find if ^=cos {in sin"^ .v), Sol. Here >-=cos (m sin-' ;«) (y)^=i or or or or >’i= — sin (m sin ' a). m (Vi)o— 0 •••(I) • 1 — sin* (m sin“' x) =hF[1-cos*(«) sin-' ,v)J=m3( Differentiating again, we get 2yiy2i I -x*) - 2x>'i2= -2mV>-, y2{i-x^)-xy^-\-in^y=0 (>'2)o+"i’“(.v)o=0 (>’2)0 — — W“(;‘)o=— m- (■•• Differentiating (1) /I limes, wc get Putting x=0, wc get l-my^ — O (.F..42)o=(«2-/«*)0-„)n Let n be odd. Putting « = 3, 5, 7, etc., we get Wo=(l^~m*)(;'i)o=0 f>'5)o=(3*— /n*)(;-3)o=0 04)0 = (5*-/m*)(_^»,,,)o=o Hence (>-«) = 0 when n is odd. Also putting /i=^4, 6, 8, etc., weget (yi)o^{2^-m^)(y2)o 0’«)o = (4*— m*)(>'4)o = ( 42 -m*K 2 *-w*)(~m 2 ), etc. Hence when n is even, we have ( j’")o= {(«— 2)2 — m*J( 2)3 = {(« - 2)* - /«*) { (/I - 4)2 - m2 j ( y,.. 22 -m^)( - /„)2 (>i)o=0 (>'a)o=0 ^>'5)0=0 (>’2)0 /W2 8g A TEXT BOOK OF ELEMENTARY CALCULUS Exercise XIII Find the nth differential coefficient of : 1. A® cos A. 2. A- log X. 3 e" log A. (K U. 1976 4. 5. atV-'. 6. aV'. 7. sin bx. 8. If y = a cos (log x)-rb sin (log .x), show' that (i) x\v, j+A-yi+y=0. (ii) x^y, , 4 2 4- ( 2 n 4- 1 ) Av,. + , -f (rtH I) v,. = 0. w 9. If y=e'* cos A. show that y 4 4-4v = 0. 10. If y=sin* A, prove that .>.d^v dv (1— A- j-t,=A "• ' d\- dx 11. If y=e^^ ■' . show that and ( 1 - x-)>'n ^2 - f 2n -r- 1 )xv« +,-(«*+ o‘>v,. = 0. Deduce the value of ( v«)o- 12. If ^’=(sin * .v)^ prove that (fi ) ( 1 — A - ) v,.4 2 — ( 2n + 1 ).v ;',.4 1 — n® r„ = 0. 13. If ;'=(a-“— D". prove that (.x®— 1 H’„+ 2 -:- 2.x/,.+i -n(n+ 1 >>^^=0. 14. {K.U. 1975) lf>’=[.v+ v'l+A**j'", show that ( i ) >’2( 1 + — /n2\' - 0. and (»f) v..42(l - -Y2)^ (2//4-l)Ay„4r; (n*-mV-*=0. {K.U. 1976) Also calculate (y„)o- 15. If t/=sin (m sin"^ .v), prove that ( 1 — A- ).v ,.+2 — ( 2/j + 1 )AyR4i — (/)*— m^)yn — 0. 16. If y'"’* +y = 2 a-, prove that (.x2-l)yn4a4-(2n+l)A-ynn r(«*-/n’*)y„=0. {K.U. 1977) 17. If v=e‘"^X. where X is a function of .x, show that >,.= e''"(X,. + "CxaX.., + "C2fl%-2-i- ) wheie Xi, Xs, etc., denote the successive derivates of X with respect to x. 18 If M=sm «A' 4-cos nv, show that i~ ir sin 2n.v}» * where u denotes the rth derivative of w with respect to x. {K.U. 1962, Supp.) Derivative as a Rate Measurer 61. In differential calculus, we are always concerned with the variation of related quantities. This means that if one quantity changes, we wish to know the rate of change of any other related quantity. This is explained below by means of some examples ; Let y=f{x) be any function of x. If h be the increase in the value of X, then /(x-j-/i)— /(x) will be the increase in the value of fix), and the ratio represents the average rate of increase in fix) as x increases from x to x+/j. If /; be taken smaller this ratio will measure more and more approximately the rate of increase of fix) relative to x at the particular value of .v under consideration. Therefore, when h -*0, Limit measures the rate of change of fix) relative to x. Hence ^ measures the rate of change of y with respect to x. <►‘2. Rate of change with respect to time If a quantity x be a function of the time t, then measures the rate of change of x with respect to the time t. *““®**’**’®“®- L«t the position of a moving point in a straight hne at timer be fixed by measuring its distance x from the origin O in the straight line. Let this distance be x+Sx after time r+ 8/. t Sr Then the total space described in 8/ o ' — Z .r\ time is 8x, and, therefore, the average r^xg speed is equal to Now as Sr -»0, the rate at which the space is described at time r is represented by This is known as the velocity at time r and is generally denoted by v. 90 A TEXT BOOK OF ELEMENTARY CALCULUS Similarly, ihe rate of change of velocity at time t will be re- presented by ^ . This is known as acceleration, and generally denoted by / Solved Examples Example 1. Find the rate of charge of the volume of a- circular cylinder of radius r and height A when the radius varies. Sol. Let V be the volume, then v=nr^h Rate of change of v w.r.t. r is dv dr = 2nrh. Example 2. A particle moves in a straight line and its distance S in feet from a Bxed point O is given by its velocity and acceleration on each occasion when it passes through O. Sol When the particle passes through O, S^O. Naturally ,,,-1)8=0 which gives /=0 and f=l. As such, we have to find velocity and acceleration when /=0, and f=l Now S=-/®-2/» • f % # v=-^ 1 (it and 11 Therefore, (0 when /=0, v-1', and and (fV) when t=l, v=0, and Note. Acceleration is denoted either by dv ils dy_ dt /--4';sec“ /=2'/secV or by or by Again, Hence dv _ — 1 (ds ] i d-s ill dr [dt j 1 dt^ dv _ dv ds dv dt \h (It ds f- dv dv _d^s '' dt = v ds dt"* — ds Example 3. A particle moves along a straight line according to the law „ .s / - \ v*=2 (sm .Y — A* cos .x) where v is the velocity and .v the distance described. Find its acceleration. DERIVATIVE AS A RATE ME.\SURER 91 Sol. v®=2 (sin X— X cos x). Differentiating w.r.t. x, we get 2v 4^ =2 (cos X— 1 . cos x+x sin x) ax f dv • • /=v -j- =x sin X. dx Exercise XIV 1 . Find the rate at which the following vary with respect to a change in radius ; (i) The area of a circle of radius r, (ij) the total surface of a cylinder of radius r and height h, {in) the volume of a sphere of radius r. 2. A point moves so that its distance S from a fixed point O at lime t is expressed by S=fle~** sin (w/-ha). Find the velocity of the point at any time t. 3. If x*+y*=25, find the rate of increase of y at the point x=3 ; also at the point x= —4. 4 . The radius of a circle is increasing uniformly at the rate of 3* per second. At what rate is the area increasing when the radius is 1 ft. ? jgj^) 5. If a body is moving in a straight line and its distance in feet from a given point in the line after t seconds is given by S=5+2/+4(«. (K.U. Old Course, November, 1976) Find (i) the speed at the end of 2J seconds, (lO the acceleration at the end of 3i seconds, and (ill) the average speed during the 4lh second. 6. A particle moves along a straight line, the law of motion being S=A cos {nt-\-k) show that the acceleration is directed to two origin and varies as the distance. 7. A point moves in a straight line so that its distance S from a fixed point at any time f is proportional to r". If v be the velocity and / the acceleration of any time /, show that 92 A TEXT BOOK OF ELEMENTARY CALCLLUS 8. A particle moves along a straight line according to the law Prove that the acceleration varies as -p • {K.U. November^ 1975) 9. The speed of a particle moving along the axis of x is given by If / is its acceleration, show that 27v*=8(2-/)(4+/)*. A triangle has two of its vertices at (—0,0) and (fl, 0) and the third (x, >») moves along the line y— mx. If A be its area, show that (K.U. November, 1976) (Hint. A=i.2om;’=mox etc.) '^1. The diameter of an expanding smoke ring at timer is proportional to If the diameter is 6 cm. after 6 seconds, at what rate is it then changing ? {K.U. 1977) Tangen ts and Normals 7*1. Definition Let P(x, y) and Q(x+Sx, be any two neighbouring points on the curve y—f(,x)y then the limit- ing position of the chord PQ as Q tends to P along the curve^ is called the tan- gent to the curve at P. 7-2. Equation of the tangent to the curve y=f(*) at P(x, y) LetP(x, y)and Q(x+Sx, y+M be any two neighbouring points on the curve y=yix). Then, as we know from Co-ordmate Geometry, the equation of the chord PQ is y+8y-y x+Sx— X (X-x) where X and Y are the current co-ordinates. or Now that the tangent at P(x , y) is the limiting position of the chord PQ as Q ♦P, pro :eeding to the limits, as Sx->0, we get the required equation as (X-x) Lt 8x ''dx ) A TEXT BOOK OF ELEMENTARY CALCULUS lly 7‘3 * Geometrical Meaning of ^ In Article 2-5. it has been sho^ that of rential co-efficient of a function fix) at some pomt ts the slope oj the tangent to the graph of the function at that pomt. Thus if the tangent PT to the curve y=f{^)^ at P(x', v') makw \ I with the ^is of x, then the slope of this tangent, t.e. tan ^ t^auautThe vaLof the differential co-efficient of the funetton a° the point P(.v', /). ^ This can be written as , /av \ )x\y' 7 4 Equation of the Normal to the curve y=f(x) at P(x, y) Definition. Normal to a curve at any point is a straight Ime through that point perpendicular to the tangent at that pomt. Hence the slope of the Ungent is that of the normal will . Hence the equation of the normal at P{.t, >■) is ’=13. Next, slope of the normal = | The equation of the normal is >— 2=Kx-3) or 2x—3y=0. 2. Find the equation of the tangent to the curve y’ = 4a.x at any point. or Sol. Let the point be P(x, y). Differentiating the equation of the curve w.r.t. x, we get dv 2a dx ~ X or or or The slope of the required tangent at P(.y, y) ^(‘±) \ dx lx, y y Hence the required equation is Y-y=— (X-x) y yY-y-=2flX-2ax yY:=2flX-2ax+y- = 2flX — 2ax+4n.v y- — 4ax) = 2a(X+.Y). An Important Note In illustration (1) we have not taken capital letters as current CO ordinates because the point of contact is given there in clear terms. In the second illustration, on the other hand, we have taken y) as the point of contact which was not given, and to avoid con- fusion, we have taken current co-ordinates in capital letters in it. Thus, current co-ordinates should be taken in capital letters where the point of contact Is not given, and where we take it to be (x, y). A TEXT BOOK OF ELEMENTARY CALCULLS 'Equations of tangent and normal when the Parametric Equations of the curve are given Suppose the equations of the curve are : ■ T =/(0 .v=9(/) Now from (i), we have and from (ii) dv dy fdx ?-'(/) dx~dtl dt ~ f\t) ' Hence the equation of the tangent at [.v=/(r), v= 9 (f)] or at “/’’is >— [.V -/(/)] Again, the slope of the normal 1 dy ' 9'(/> dx Hence the equation of the normal is or [-v— /(nj/'f/l-i-Lv— 9(/)J 9'(/) = 0 / Note. As is obvious above, we can easily avoid taking current co-ordinates in capital letters in case the equation of a curve is given in parametric form. Illustration. Find the equations of tangent and normal at the point “0” to the ellipse x=a cos 0, y—b sin 0. {P.U. 1965) Sol. x=a cos 0 t • dx dO = — sin 0 and y~b sin 0 ^ = 6 cos e. dy___dy jdj^ b cos 0 dx rfO ' tfO ~ a sin 0 Now TANGENTS AND NORNULS 97 or or Hence the equation of tangent at (o cos 6, b sin 6) is . . h cos 6 , a\ y—b sm 0= — — : — ;■ {x—a cos 6) o sm U bx cos 0+ay sin Q=ab (cos^ O+sin^ b)=ab X cos 0 . >» sin 0 , fl b or or Again, the slope of the normal _ a sin 0 ” b cos 0 Hence the equation of the normal at {a cos 9, b sin 9) y~b sin 0=-r z (a—o cos 0) 6 cos 0 ax sin ^—by cos 0— (a-— 6®) sin 6 cos 0 ax sec Q—by cosec 9= (a*— 6*). Solved Examples Example 1. Find the equation of the tangent at any point to the curve y— f(x ) . {K U. 1 950) Sol. Let (x, y) be the point of contact. Now the equation of the curve is y=fix). Hence the equation of the tangent at (.v, >») is Y-y^fix)[X-x]. Note. The above example is extremely important and can be taken as an article. E xam ple 2. Find the equation of the tangent and the normal at the point “r” to the curve whose equations are > = flsin*f. (K.U. 1964, Sov., 1975, 77) x=a cos'* /, From x=a cos'* t, we have ^nd from dx dt = — 3a cos* / sin / y~a sin* /, dy -r~‘5a sin* / cos / dt dx~ dt dt 3a sin* r cos_( 3a cos* t sin t sin t cos t 98 A TEXT BOOK OF ELEMENTARY CALCULUS Hence the equations of the tangent at (a cos® a sin® /) is sin t y—a sin® /= — cos t (x—a cos* /) or or or y cos t—a sin® / cos t=—x sin /+n cos® t sin t X sin t-^-y cos t=a sin t cos / X sec t-\-y cosec t=a. Again, the slope of the normal JL dy dx cos t sin t or or Hence the equation of the normal at {a cos® t, a sin* /) is sin® /= , V sin t—a sin* t=x cos t—a cos* / ,3 S^LL {x—a cos® /) X cos t—y sin t—a (cos* t sin* 0 (cos* /—sin* /)x(cos® f+sin* /) or X cos t—y sin cos It. Example 3. Obtain the equations of the normals to the cur^ 6 )=a:® at the point, where y=7. {K.U. Nov. 1976} Sol. The equation of the curve when simplified is Differentiating this with respect to x. we get dy 2x or dx "■ Now >'=7. therefore, substituting this in the equation.of the curve, wet get, x=±2. The points at which we have to find the eqations of the normals are (2, 7) and (—2, 7). 1^°*' )<2, 7,= -^ [ix )(-2, 7)=-y Slopes of the normals are —1 and i. Hence the equations of the normals are y—l= — \ (.V— 2) or 5;c+4^=38 and V— 7=5(j:+2) or 5.x-4j'+38=0. Example 4. Find the points on the curve _j,= 3x«-6x+5 where the tangents to the curve are parallel to the axis of x. TANGENTS AND NORMALS 99 Sol. The slope of the tangent which is parallel to the x-axis IS zero* Differentiating the equation of the curve w.r.t. x, we get dy dx — 6x—6. As the tangent is parallel to the axis of x, we have dv ^=0 or 6x— 6=0 which gives x= 1. Substituting this value of x in the equation of the curve, we get.y=2. Hence the required point is (1, 2). Example 5. Find the point on the curve at which the normal makes an angle B with x-axis. Sol. Equation of the curve is Differentiating w.r.t. x, we get dx dx Slope of the normal -1 3 -1/3 A '8 aTs jly dx X/3 ¥ , 1/8 By the given condition, .1/3 T7r=tan 0 • • ft ( 2 ) .1/3 83/3 — -_£!!!±2!L!L — = Q .2 /3 by{l) and sin* 0 cos*0 sin*0+cos*O x^'^=a^'^ sin^ a or x=asm=’0 cos^o or y=a cos® 0 the point is (a sin® 0, a cos* <>). Exercise XV *'"= ‘“S«nt and the normal atony point to lowing curves : ^ (/>r^=x*. (w^+/= W. (iv) x=at\y^2at. (v) x=a sec p~b tan (vi/ x=a (r+sin 0, y=a (I —cos /). {K.U. Nov. 1976) JOO A TEXT BOOK OF ELIMENTARY CALCULUS 2. Find the equations of tangents and normals to the following curves at the points indicated against each : (/) at {x\ y) and (a cos 6, b sin 9). (i7) xy=c^ at (x'. /). (in) 3/=^ at (3, 3). (nO (v)y^y= 2 ax at (2fl, 0). ^ SUow that the tangent to the curve y=3x*-4x^-\-l is parallei x-axis at the point, where x— 1. Find the points on the curve >»=12x— X* at which tangents are parallel to the axis ot x. 5. Show that - +-f = 1 touches the curve at the point where it crosses the v-axis. 1976) 6. Show that for all values of n a 0 touches the curve ( --) +( y ) =2 at the point (a. ft). {K.U. Old Course, 1976) 7. Find the equation of the tangent to the parabola v-=4x+5 parallel to the line 2x-r=3. . . ^ 8. Find the equation of the normal to the curve 3.\" y —14 parallel* t^-!-?r=^* 9/ The tangent to the curve v' v+ v'y=y/o meets the axes ot V and V at P and Q respectively, show that OP+OQ=fl. 10 Show that the portion of the tangent to the curve which is intercepted between the axes is of constant length. [Hint. Take (x, y) as the point of contact.] (K.O., B.A., 1977) 11, Find the coujition that the line /x+z/iy -!-«=0 may touch the ellipse “o ' V 3 ” 1 • ly Show that the condition thht the line -x cos a ->» sin a— p may'touch the curve is m+n = rt)"*"" . a-^'- . cos”* o sin’* a. ^ {K.V.B.A. {Pr.) 1977) TANGENTS AND NORMALS 101 r5; Find the condition that the line X cos a + v sin '=/(x) and >»= /(x) intersecting at a point P(Xi, >'i). Then the angle between the respective tangents to the curves at the point P is called the angle of intersection of the two curves. 7*7 The following is the working rule for finding thcangle between two curves First step. Solve the equations of tlie two curves simulta- neously in order to find the common point oJ intersection. Second step. Find from the equations of the two curves separately. Third step. Find the values of — at the point of inter- section found in the first step. Denote these values of A and m 2 . by mj 102 A TEXT BOOK OF ELEMENTARY CALCULUS then Fourth step. Use the formula tan l+Wimg where 9 is the angle between the two curves. Note, The supplement of this angle is tan-^ r~ 1 +mim2 Cor. 1. If the two curves cut orfhogonaliy at P(xj, yj) then / '{xi).9'( A'l)— — 1 • Cor. 2. If the two curves y=f(x) and >'= 9 (x) touch at P(^i. yi) then /'(xi)=9'(xi). 7 8. To show that the curves ax‘-{-by*=\, and /x^+my^=l cut at right angles^ if 1111 a ^ b ~ t m Let P(xi, >’i) be the point of intersection of the given curves, then n.vi»+h>’i2=l and ix^-k-my^^\. Solving these equations for X|* and >^ 1 *, we get .> m — b , 8__£ZiZ_ ~am -br am~bl From the equation of the first curve, we get dy ox dx ” by Similarly, from the equation of the second curve dy_^_i^, dx my Therefore, the slopes of the tangents to the two curves at P(-Xi, ^i) are respectively nxi . Ixi — i— ‘ and • by^ my^ The two tangents will be at right angles, if bm}Y i.e, if o/.V|2=— hmyi®. Now substituting the values of Xj® and >h*, we get al{m—b) bm{a—l) am — bl am~bt 1111 (mima— — 1) TANGENTS AND NORMALS 103 Solved Examples Example 1 . Find the angle of intersection of the parabola >^ = 2 x and the circle Sol. Solving the two equations simultaneously, we get (2, 2) and ( 2 , — 2 ) as the two points of intersection. or and or Now from = 2 v ^ = >^= 2 x. dx Also from ;c*+>’*= 8 , “ dy y (*)(2 ,2) =-!=">. (say) Hence tan 9= = 3 l+Wimj l + l(— 1) 0=tan“* 3. Again, from equation (D- ^ at (2, — 2)= — i=mi dy 9 3 >> (2). ^ at (2, -2) = l=m, e' = tan-H-3). Example 2. Show that the two curves .v®— 3.xy*=2a* and y®— 3xV=llfl* cut orthogonally at (2^= 0 ^:, x-=by. 7\S: y*^ax, x^-\-)^=2axy./0 7. Show that the curves .>^-3x>>*+2=0 and 3x^y-)^^2 ^ cut orthogonally. 8. The curves xy=Ci and pass through the point (3, 4). Find the angle at which they intersect. 9. Show that the parabolas and y^^laji intersect on the curve x^+^^^oxvi^lso find the angles betwe’en each Tpoint at the points- of intersecticm. piXvv 10. Show that the curves intersect at right angles. [Hint. Proceed as in 7'8.] 7 - 9 , Lengths of Tangent, Noniinl> Snbtan^ent and Subnormal Let PT be the tangent to the curve y=Ax) at any point P(.\-, y) making angle 4* with the positive side of the x-axis and let PT' be the normal to the curve meeting the .v-axis at T'. Draw PT X upon .v-axis, so that PL=y. Then we have the following definitions : 1. Pt, the por- tion of the tangent in- tercepted between the point of contact and the axis of .v, Is defined as the length of tangent. 2. PT', the portion of the normal intercepted between the point of contact and the axis of x, is defined as the length of the normal. TANGENTS AND NORMALS 105 3. TL, the projection of PT on ;c“axls, is dehned as the length of the subtangent. 4. T'L, the projection of PT' on x-axis, is defined as the length of the sub-normal. Now ’. Therefore : 1. Length of the tangent=PT=>> cosec >>=1/ l+COt* =. V *+??- 2. Length of the normal = PT' sec =y- lTun^=>' 3. Length of the subtangent = TL —y cot i> y Xi 4. Length of the subnormal = T'L=/ lau v=xvi- Solved Examples Example 1. Show that in any Cartesian curve 1 + v,^ r Tangent " L Normal _ Subtangent Subnormal (K.U. Inter., 1957) Sol. Tangent= ^ v H-y,* Xi Normal= V 1 +yi^ Subtangent Xi and Subnormal Now L.H.S. ~ Tangent “j Normal J JLy/l^y2 Xl [ X^l-\-y,^ J 1 106 A TEXT BOOK OF ELEMENTARY CALCULUS y R „ _ Subtangent j', 1 .n.5).— 773 >’>'1 )\ Subnormal Hence L.H.S. = R.H.S. Example 2. Show that for the curve y=b^'* the subtangent is of constant length, and the subnormal varies as the square of the ordinate. Sol. Here = dy b ... 1 = A dx c y=be*f) y y Subtangent=^ =c (constant). dx Again, subnormal Subnormal _ dy ^ ^ y y^ ^ dx ^ ' c ~ c V' = — (constant). c Hence the subnormal varies as the square of the ordinate. Elxercise XVII 1. Show that in any cartesian curve : (/) Subtangentx subnormal — (ordinate)*. (//) (NormaI)*=(subnormal)2H- (ordinate)*. X 2. Show that in the curve y=6c " the subnormal varies as the square of ordinate, and that in the curve y—ef, the subtangent is constant. 3. Show that the tangent to the curve Ar+ =log V a*— y* T iK'M T is constant and equal to a. 4 . Show that in the curve .v— V b^—y^—b log {6+ V b-—y*} the sum of the subnormal and subtangent is constant. 5. In the curve v v\ ^ x=fl(cos r+log tan i/), y—a sin r ^ ' prove that the portion of the tangent intercepted between the x-axis and the curve is of constant length. 6. Find the lengths of tangent, normal, subtangent and sub* normal at the point 9 on the (^rve x==a cos* 9,^=a sin* 0. {K.V. 1975) TANGENTS AND NORMALS 107 point Of the subtangent at any varies as the product of the corresponding co-ordinates. 8. Show that the subtangent at any point of curve varies as the abscissa. Show that the subnormal at any point of the curve f-x^=aHx^-a^') varies inversely as the cube of its abscissa. 10. Prove that for the ellipse — + -^ = 1 2 r 1.2 a length of the normal varies inversely as the perpendicular from the origin upon the tangent. {P.U.) 11. Prove that for the catenary >'=c cosh the perpendi- cular dropped Irora the foot of the ordinate upon the tangent is of constant length. 12. For the curve y=a cosh ^ j, prove that the length of the portion of the normal intercepted between the curve and the iKU. B.A., 1974) \ H. ^ I- (^a .1 •x-axis is , V 4A Integration 8’1. Integration as the reverse of differentiation Integration is the process by which we can find a function whose dijj'erential coefficient is given. It is thus a process which is the inverse of differentiation. For instance, let us take the case of x*. If we differentiate x" w.r.t. n we get nx’*-^. Now nx’*'^ is called the differentiate co-eflBcient of x” w.r.t. x, whereas .v” is the integral of f?x"-^ The process of differentiating x” with respect to x is denoted by the equation 4 - (x")=nx"-' dx while the one of integrating nx’‘~^ with respect to x is denoted by the equation dx^x’*. Thus, the symbol /r/x denotes the process of integration with respect to x just as the symbol ^ denote the process of differentia- tion with respect to x. 8’2. In the preceeding Article we have defined integration as the inverse process of differentiation. Consequently, if we are required to integrate cos x w.r.t. x, we are evidently required to find out a function whose differential co-oefBcint w.r.t. x is cos x. Obviously, this function is sin x. This can be symbolically represen- ted as /cos X dx = sin X and is read as ; “integral of cos x w.r.t. .Y = sin .x”. 8'3. We give below some of the elementary fundamental formulae of integration and the student is advised to commit these to memory. Most of these have been proved in the forthcoming articles. I. Algebraic Functions 1 x" dx= x"+i rt-f- I 1 . (provided n#— 1> INTEGRATION 109 3. 4. 5. 6 . 1 It ='°® •" — log (flX+i). ]ax-hb a 1 1 (provided — 1) dx dx 1 ^ .1 X = — tan ' — a a Va^—x^ =sin”* 2 7. [ sec-i J-x^'x^-aa o a II ^ Trigonometric Functions 1. Jsinx(/x=-cosx. 2. fcos x ^/x=sin x. 3. /sec* X rfx=tan X. 4. fcosec^ x rfx=-cot x., 5. Jsec X tan x dx^szc x. 6. Jcosec x cot x dx= -coscc x. 7. /tan X dx = \os, sec x. 8. /cot x dx = \og sin x. III. Exponential Functions f e*"* -» 1. e’"* dx = . 2 J m S*4. Two important formulae - K j a"*' rfx= .mx m log a. '(x) fix) dx-log/(x). 1 . Proof : 1 _ [ /(x)]V'UK«+0 _r i/xL «+i -I '*+1 r f Ax)l"+^ Hence j[/(xj]V'(^W-<= „+ r Hence ^/x=log/(x). S-5. General Rules for Integration Theorem 1. A constant factor can be taken out of the integral sign without affecting the value of the integral. Or, symbolically, / af{x) dx-a I fix) dx ••■( 1 ) 110 A TEXT BOOK OF ELEMENTARY CALCULL'S Proof. Differentiating both sides of (I) w.r.t. x, we have <+i Cor, x" dx= X n+\ This is a special case of formulae I and can be proved b> differentiating the R.H.S. It ““>8 [These can be proved by the same way as formula (1).] Illustrations Cor. (Illustrations on formula /) (3x+4)‘. j(2x+5)-^'2f/x . (2x+5)-i'2*i_, i2x-\-5)^'^ -i=H 4 = V2x+5. (/) J (3x+6)=‘c :.v) \ V2X-1-5 1 (8-9x)-«^^ 1 1 - 9 -3 + 1 "18 •(8-9x)*‘ 1 1 3 • X® ■ (Illustrations on formula II) (i) Hi) ii^x=-T '“6 'OB fTr. ^ Please note that —log x=log ^ J 112 A TEXT BOOK C»F ELEMENTARY CALCULUS in. (a) f sin* X tJx=i(x—i sin 2x), 1— cos2x Proof. sm^ f s j f 1 — cos 2x , I sin* X dx— J 2 ^ =iU— i sin 2x). ni. (6) /cos® X dx=^{x-\-\ sin 2,v). The proof of this formula is left as an exercise for the student. IV. {a) /tan® x dx=tan x—x. Proof, /tan* x d.v= /(sec* x— I) dx =/sec* X dx~ldx=X^n x—x. IV. (6) /cot* X dx= —cot x—x. The proof is left as an exercise for the student. V. (a) /sin* x dx— i^i cos 3 .y- | cos x. Proof. We have sin 3x=3 sin x—4 sin® x sin* .Y=i sin x— i sin 3x Hence /sin* .v d.Y=/(J sin x— J sin 3x) dx = J /sin X dx— i /sin 3x 1 , 3 cos 3x j- cos X. 2 4 V. (i>) /cos* .V dx= i ^2 sin 3x+ J sin x. The proof is left as an exercise for the student. Note. Other fundamental formulae will be proved at their proper places. 8’7. Constant of Integration We know that and Also rfT <''^+3) = 2-v ^ (.v*+c) = 2.y \* ^2x dx=.\- |2.y dx=.Y* + 3 J2x dx=x*+c. This shows that X*, .Y* + 3 and x*— c are the integrals of the same function 2x. Hence it is necessary to add an arbitrary constant c to the result of integration. INTEGRATION 113 Therefore, if ^ /W=9W then h{x)dx=f{x)-\-c. This constant “c” is called the constant of integration. Though it is frequently omitted from the result, its presence is always understood. ^'8. Definite Integrals Definition. If is a function such that Tx Then the definite integral of fix) from a to b--=/^-2a(a-}-xV'^ \t^. {K.U.1977) Jl + sin.v ^ Example 4. Sol. We have f dx __ f d— sin .v) dx Jl + sin.v ] 1 — sin" -v (1 — sin x) dx COS'* .V =1 Jcos^.v Jcos" X =/sec- .Y Jsec x tan x dx — tan A‘— sec x. INTEGRATION 115 Example 5. / v r-rJ>m .v dx Sol J v' l + sin X dx ** I ** V* I ^ y 4-sin' ^ + 2 sm ^ - j cos^ = j/^^cos y + stnyj dx = J^cos Y +sin X cos dx sin^ cos^ ^ . X X = 2 I sin COS i i 2 — 2 )■ Elxample 6. |sin 3a: sin x dx. Sol. sin 3 a: sin x=-\0. sin 3^: sin x) = i(cos 2x— cos 4a:) /sin 3x sin x ^64) 116 A TEXT BOOK OF ELEMENTARY CALCULUS 3. Evaluate : iis «> iKf; IsTT «» m <« 17 =f <«J^ 4. Evaluate ; dx. dx. (.) ^yUO J Vx +3 (x) [ X V x+3 dx. X dx Vx*+3 (i7) /sin {>^+x) dx. U) jcos -y dx. Hi) /sin {»t+x (Hi) I cos jrfx. (iv) /sec^ x dx. (v) /sec^ 4.V dx. (vi) f ■ ■ Jcos^ ax (vii) [ tan- .V c/x. (viii) | cos^ x rfx. Hx) f Csin A— cos a)- dx. (a) [ - . — 5 — J J sin- A COS'* X fHint. ^-1 L sin- A COS- X sin- a cos- a J (vi) [ in) I (viii) 1 cos^ X dx. i/O COS 2 'J (xii) j (tan^ A~cot“ a) (/x. ..xiU) I sin 3a cos 2a dx. (a/v) j cos 2a cos 3 a dx. (At/. (at) I cos px sin qx dx . (p>q). (At/. J964) (XV/) • Jl-rCO<; A (XV///) ./-v. (xx) j ^ I +COS A dx. I - v f dx [w X , — sec^ Y X X sm Y" I .. X . 'Ysec- “Y "X X X sin 2 ' cos Y 1 -sec dx X tan — Putting tan y we have ^ sec"'^ Y^x=du 1 2 124 A TEXT BOOK OF ELEMENTARY CALCULUS 3. ^ Questions of the type ^ 4 . (0 f (« 1 dx JxVx»-2 xV9x*— 1 («7) f ^ (iv) 1 dx J2xx/4x2-l tVt-' ^Questions of the type IVig *) !VS «> 1 V 1-i (m) X dx. (Hint. Put cos 26> 5. ^Questions of the type j ) {»0 j (1+x*) tan-' .V dx J .Y log a: , ... r 1 — sin .Y (v») )- Y + COS .Y sin X dx dx. (n) f 3 x't- 1 J3x*+2x+I {(v) f sec* X . 1 : dx. J tan X iyi) (vm) f cosec* X dx col X -\-b cos -Y (tan"' x) (nil Q ixiiol ^ (.YV) j «/2 COS .Y dx dx. -f-sin .Y U) JVi-.t 2 sin-1 X r ... f COS A 1 TTii (xiv) J- sm* X dx Y(i-rlog x) iK.U. 1975) sm X cos X dx. 3 COS" x+2 sin" x 6. (Questions of miscellaneous type) J .tan"! V {K.U. 1975) •>i (i7) 1 ^ cos ^ dx. INTEGRATION 125 X)2 ..... f (tan'^ . <»0 j-TTP (v/)- 1 sec’ X t dx. ( \ 1 . \i tan x \ - v<.»+ 1 i V— )• r ^+I INTEGRATION 127 Example 4 I J (o si sin x+b cos .x)’^ get Sol. Dividing the numerator and the denominator by cos- x, (Ix I (a si dx cos'^ X sin x+b cos x)- ( ^ cos X / -1 sec“ X dx (a tan x+6)- Pulting tan x=m, we get sec^ X dx=du sec- X dx ^ ® J iau- {a tan x-\-b) — du= br- {au+bi ' 1 1 — a a au I 1 a a tan x+b Sol f dx Jxv X*— I f dx _ f X dx Example 5. dx (/(.U. 1976) 1 Put x*=scc 6, so that x dx=\ sec 0 tan 6 d9 . f X dx _ [ h se c 0 tan 0 j f sec tan 0 f/ 0 J-*^Vx*— l”^Jsec 0 Vsec“ tt— J sectflanU = i / tl = J sec'* x-. Exercise XX Evaluate the following : dx 1 3 . 1.1 I dx sm x+cos X d^ , 5 cos X— 12 sin x I ■ 1 - 3 sin x+4 cos x sin X dx {K.U. 1974) 5. (/) 6. i) J V x^-hfl- f dx J 1 + 3 sin“ X* sin (X — a)' [Hint : sin x=sin (x — a-ra)] ov) J V X^-fl- 7. f ' - ^ — dx. Jo VI— X* 128 A TEXT BOOK OF ELEMENTARY CALCULUS 8 . Jl+cos*x , ^ r • Jo 2fsin*x* • J 5+4 cos 28 li. f«/2 dx 12. Jo 1+4 sin* x’ 13. . /I'TT t077\ 14. Vl+x* 15. r dx Jsin (x—a) sin (x— 6)* 16. dx sin x+cos x)* 14. 1^/0®— X* dx. dx 3+2 sin x+cos x X 2 tan Hint. Put sin x= and 1+tan 2 COS X* T 1+tan* -J- 17. 19. 20 . f sin X dx J V 1 +sm x’ io r sin 2x dx * J(a+ftcosx)** f sec X log (sec x+tan x) dx. c log X dx. [Hint. Putcosx=r] {K.U., B.A., 1966) 8* 13, Integration by Partial Fraction Sometimes the expression to be integrated can easily tesolved into partial fractions, which facilitates the integration. This •is illustrated below by means of a few solved examples : Example 1. I Sol. Let Bx+C ■x— 2 _ A (x+lK.x*+l) '"x-l“^ .v*+l x-2 = A(x*+1)+(Bx+CK.v+1). Pulling x= — l,weget 2A=-3 A=-t. Also, equaling ihe co-efficients of x* on both sides, we get A+B=0 which gives B=|. Also, equating the coefficients of x, we get B+C = l which gives c=— + . x-2 . -g_ Ix,^. (x+l)(x*+l)"* x+1 ^x«+l INTEGRATION 129 Hence ! x -1 I X dx , f , x2-hl * Ja-2 ^ [ dx , \ 2 x (ix , r < + 1 dx + 1 = — ! log (X+l) + J log (x2+l) — i tan”^ X. Example 2 dx 1 dx xU’*-i-l> Sol f dx f x’^-W/v • Jx(x’* + 1) Jx"(x'‘+1) (X t/.. B./4. /P57) (Please note this step) f x” ^ dx Jx"(x-+1) JL n J_ n ]_ n _l_ n _L /] 1 /7 1 dz 1) \_ n n z-\-i .Y'‘ x"+l Example 3. Show that dx I W \^ 2 = 2; , x — a 2« (x a). (0 \ (x-\-a){x~a) 1 f(_L„__LU 2tfJ'x — rt x-fn/ = ^(log (x — a) — log (x-r (. * 1 , x-a 130 A TEXT BOOK OF ELEMENTARY CALCULUS (") o»‘^x»~I(a+J^o-x)“2fl Ka+jc+a-x) = ;^tlog (a+Jc) — log (fl— x)l 2a 1 , fl+X Note. Example (3) is of great importance and can be treated as an article. The student is, therefore, required to commit it to memory along with its result. Example 4, Evaluate (!+,;„ x)(2+sin - x) ' Sol. Pul sin x=z, so that cos x dx=dz. Also when x=0, z=0; when x=-r-, z— 1. ■jt/2 cos x dx Jo (1+sin x)(2+sin x) -i: =1 frl 'o Example 5. Evaluate =i: dz (H-2)(2+z) (Please note tliis step) 2 , 1,4 -log -:r=log 3 dx 5 sin* X iK.U. 1976) cosec* X dx cosec* X— 5 Sol. f . , =U J4— 5sin*x J4 I CO sec* X dx 4(l+col> x)-5 I cosec* X dx 4 cot* X— 1 Put cot x=i so that cosec* x dx—-~dt (Please note this step) The integral _ f f dt Jr=^~ J(l + 2f)(l-2f) + ~ 2 l(n-2/"*'l-2r)^' = log (l-f2f)-* log (1-2/JJ 1 . 1+2/ 1 , 1 + 2 tan X = T r^f=T 1-2 tan x* INTEGRATION 131 Example 6, Evaluate r Sol. Put so that e* r/jc= dt i.e.. e* /+! *• the integral =j ,'2 dL (/+!) (K.U., B.A. 1977) or Now, let _A, B C r“(/+l) t 1 I=-Af (i4-l)-rB(r+l)+C/- Putting /=0, and /=-l, we get B=1 and C=l. Again, equating co-efficients of t- from both sides, we get A+C=l or A=-C = -l \i4+T)^\{-'T + 'h+7h)‘‘‘ = -Iog t — ^+log (r+1) Exercise XXI Evaluate the following : 1. f dx ix^-4 2. [ dx j6x"+5x-4 3. f x^dx JU+1)3(X + 2J 4. f (x2[-6)V/x J(x=*+4)U- v9)’ 5. f x^H-l Jx(x»-1) fx^ dx 7. r 2x- dx 8. f 3x+l jU-l)(j:“+l) J(x-l)V+3) 9. 10. [ X2 dx Jjr*4-5x^-h7x+3 ^ ' J{x-)-aj{x+6j(.ri-c)] 11. f Je'-he*** 12. sech X dx. 13. f . Jsin x-hsin 2x 132 A TEXT BOOK OF ELEMENTARY CALCULUS • 813. (a) Integration by parts Rule, This rule follows from the rule for the differentiation of a product of two functions. If u and v be two functions, then d(uv)—udv-\-vdu or udv—d(uv)^vdu Integrating both sides, we get Judv= uv — Jvdu which gives the rule. Example 1. Evaluate : /xe* dx. Sol. Take m=x and v=e* Jxe* dx— xe*— Jl.e“ dx=xe*— e* . Note. Proper choice of u and v is essential as a wrong choice of these may make the integral complicated. Example 2, Evaluate : jeos x log sin x dx. {K.U B.A. Sol. Take u=»cosx and v=Iog sin x /cos X log sin X dx=sin x log sin x— Jsin x ^?^dx sin X =sin X log sin x— sin x. Note, Inverse trigonometric functions and logarithmic functions should never be taken as u. Example 3. Evaluate : /tan’* x dx. Sol. Jtan’* X (/x=/l . tan"* x dx Take u~ 1 and v •= tan * x (Please note this step) /tan"* X dx=x tan’ = x tan +x^ 1 [Ixdx dx 2 Jl+x^ =x tan’* x~i log (1+x*). V dx. 1+x (KU.yB.A 1972) Sol. Put x=cos 0 so that dx= — sin 0 d^. ! 1— cos 0 ( The integral =jtan-* yy (-sin 0) = Jtan-*^tan Y)(“Sin0)d6 = — 4 /0 sin 0 JO Jd INTEGRATION 133 = — ^ [0 (—cos 6) — /I . (—cos 6) d^] (Integrating by parts) = i cos — 2 fcos 0 = 1 6 cos ^ — i sm 0 = i . A* COS“^ A— i ) — A“. ('.* 0 = cos"' A" and sin O--'/ 1— x'-) Examples. Evaluate |sin“’ /y/ dx. {K.V.1976) Sol, Put A = fl tan- ft, so that dx—2a tan 6 sec- ft d '. the integral fsin'* a / ^ — x 2a tan 0 sec® 6 dQ } a{l-rtan-'J) = Jsin'^ sin Ox 2a tan ‘ scc^ 0 r/6 = 2aJ'0 tan 0 see® 0 d'i Integrating by parts, taking «=tan 0 sec® 0 and v=0, we get the integral =2a j 1 . ^0 ] ( Jtan0sec®0i/0 = ‘-^^j = a(0 tan® 0— /tan® 6 dH) =a[0tan®O-J(sec®0-l)i/O] = a(0 tan® 0— tan O-j-O] ={f V-s] =-(x+a) tan'* ^ ~ ^ Example 6. Evaluate |e* ^ ^ ^ {K.U. 1976) Sol. The integral=J^ e‘ i/x + log x dx. Let I — Je^ log X dx^-e^ log ■ — dx X (Integrating by parts) The integral dx+e" \og dx (Substituting for /e* log x) log X. 134 A TEXT BOOK OF ELEMENTARY CALCULUS or or Example 7. Evaluate dx. {KV. 1977) ___ Sol. The integral _ r e' dx _ f e’ dx J(A-f-l) J(x-f D* Let _ f c’ dx ~](x+l) (integrating by parts) _ x+1 Examples. Integrate \ fl®— A' by parts with respect to x. (K.U., B.A. 1957) 1= j Va- — X® dx= I 1 • ^ X* dx =xVa^-x^- ■ dx J \ a — X- (Integrating by parts) 7^1- dx J Vfl-— X* Sol. Let =.vV<,=-.v=-f dx V a*— X* (Please mark this step) — X V fl* -X 2 x*iic+a2 f — === J J V fl*— X* =xVa^— x"-l+fl' sin-» — (Art. 8TI> 2I=xV a-*— x“+fl“ sin“* t^ x\/a~— X* A J A , X sin’* — • a we get Example 9. Evaluate J e'** sin (6x+c) (/x. Sol. Let l=/e‘’' sin (hx+c) t/x. Take w=e®*‘, v=sin (Ax+e). and then integrating by parts, I=— sin (Ax fc)— f — X A cos (Ax+c) dx a i a INTEGRATION 135 or . fj V =“ sin (^-x+c)— — cos (6 a:+c) dx , A r-ai =— sin {bx+c)~—\ — cos(Aa+c) — ^ S'" J [Integrating jcH-c) — 6 cos (6x+c)l This^result can be put in another form by o=r cos 0, b=r sin 0 so that r=s/^^2 0 = t^n-i i_. a 1 = . r {sin (i^x+c) cos 0-cos (6x+c) sin 0} = ?!1lZ±^ sin (ix+c-9) a* + 6 * .ac Example 10. Show that sin ^/)X+c— tan"' V ' cos f j e"' cos ibx+c) dx= ix+c— tan”' ) This solution is left as an exercise for the student as he has to proceed in exactly the same way as in the case of example 9. Note. Examples (9) and {10) can be regarded as articles of irnmense importance, and the student is, therefore, required to memorize their solutions. A TEXT BOOK OF ELEMENTARY CALCULUS 136 Example 11. Evaluate fcos 26 log J ® cos 6— sm 6 ^ I • Sol, Take u=cos 2^ The integral _sin 26 COS 6— sin 0 , cos 6+sin 0 c -^T e-sin e - The integral ^sin 26 cos 6+sin 6 f^n 26 J cos 0+sin 01 2 ® cos 6 — sin 0 J 2 ‘ rf0 j cos 0— sin 0J _ sin 26 cos Q+sin 6 2 cos 0— sin 0 dQ f sin 26 f — sin 6+eos 6 —sin 6— cos 0 ) J 2 1 cos 0+sin 0 cos 0— sin 0 J sin 26 cos 6+ sin 6 2 ® cos 6— sin 6 fl— sin « f sin 20 f (cos 6 — sin 0)^+(c os 0+sin 6)* J 2 I cos* 6— sin* 0 4 cos*'0-sin*0 sin 26 cos 6+ s in 6 fsin 28 2 (cos* 04 -sin* 0) '> ^ cos 0-sin 0 J 2 ' cos* 0 — sin* 6' 26 sin 20 . cos 6 4- sin 6 2 cos «— sin 6 0 f sin 26 .. cos 0— sin 6 Jcos 20 ^ sin 26 , cos 04 *sin 6 — ^ >Og — -- L A 1 Example 12. Evaluate Jlog (14- x*) dx. Sol, The integral = /I . log (14-.X*) dx (Please note this step) Take u=l, r = log (I+x*) Integrating by parts, we get I 1 . log (1 +.V*) dx=x log (14-x*)— j X . dx = .vlog(l+^>)-2 =.rlog(l+.v’)-2|( (Please note this step) = x log (H-.Y*)-2 |rfx+2 =x log (I-f a;J)-2.v+ 2 tan * x. integration 137 Exercise XXll Integrate the following by parts : 1. 2. X log X. 3, X ; sin X. 4. -X sec® X. 5. X sec X tan ; Y. 6. log X. 7. x^ sin"^ .Y. 8, X® tan'* X. 9. X sin X cos x. 10. sin X log cos A*. 11. (log x)®. 12. sin“^ X. 13. X sinh X. « 14. V fl®+X®. 15. 16. sin 4x. 17. e"" tan-^ (<^). {K.U. Nov, J97o) 18. COt^ A. {K.U. Nov. J975) [Hint, col* x= (cosec® . X— I) col® X, etc.] 19 X 20 X sin'* X (fa'll IQ77\ l+A- V 1— X' •C/ . J ✓ / / / 21. . 1 —sin .X e" , 1 —cos X 22. cosec® X. (K.U. 1977) 23. €' (a* sin x+cos A ■). (K.U. 1976) Show that ; 24. 1 cos (log x) cJx = 1 ( v2 V logx- — )• 25. f.Y® sin (a log x) dx = — sin log . ,v-lan ‘ [Hint. Put x=e\] 26. ro cos 2'* 2 “ sin 2^“1 4 J where x = 2a cos 0. 27. [ cos f log — ) dx- — ^ — cos (^b log —tan"* b y J V « / \ V6®-i 1 o / fnint Put — -eO L J 28 cin Or /iv ' <;in 1 2r — tin"' ^ y mV, 1 m ^|l| fcA CiX J 'l-i-(log2)=* • dl M 1 log 2/ [Hint. T '=e' '<’3 etc.] 29 1 COS"* — dx=x J -V COS"* log (x+ V X- X '-1). 138 A TEXT BOOK OF ELEMENTARY CALCULUS 30. If u=fe'’^ cos dx dx and v=/e'’" sin bx dx, prove that (/) tan-* — +tan-* —=bx u a (ii) (a^-\-b^)(u^-\-v~)=e^^^. 8*14. Integration as the limit of a sum differenH-^^Ln*' ’’h''® integration as the inverse of Sf," L/lr./ be in the present article that a inleg^ral can also be represented as the limil of the sum of a 1 nS^'^^d*e^w '"f'^ben the number of such terms tends to innnity and each term tends to zero. 815. Fundamental Theorem of Integral Calculus , (a, b) be divided into n equal parts and let the length of each part be h, ,o that b~a=nh. then ^ " L f/(fl)+y?<3+A)+/(fl+2/;)+ 4-/(o+„':ri"A)J when ^->0 and b—a = nh. book theorem is beyond the scope of this nerformln/'l^ help of this theorem, we are now in a position to regarding it as the limit of the summation of methiS t nuinberof terms becomes infinite. The method is explained below with the help of a few solved examples : Example 1. Evaluate .x® dx as the limit of the sum. {K.U. 1977) Sol. Here yifjc)=.Y3 '■ I! -^^*'^^'’^=Ltyita=+(a+/i)2+(fl+2/i)*+ +{fl+^ h)*] Where and 6— (n-D n(2n-l) j 2 INTEGRATION 139 (Please note this step) = Lt ^{6— a) a^-\-aib—a){b—a—h) , {b~a~hMb-a)(lb-2a-h) "I ■ ^ 6 J =(6-a)a--l-a(fr— fl)2+ ( •.■ /,-*o) b^-a^ 3 * Example 2. Evaluate J dx as the limit of the sum. Sol. Here/(x)=e* e"ifx=Lt/i . , A J when nh=b - a { W-] (V e">I) /,^o ^-1 h (Please not this step) e“(c'''“— 1) Lt //-►O * (-.■ nh=b-a) =e‘' (c*’-“-l)=e''-e^ ^ VLt^^T^ l) h-*^0 h / Example 3. Evaluate : cos x dx as the limit of a sum. Sol. Here /(x)=cos X •** cos X dx=\A.h [cos fl+cos (o+/t)-fcos ia-\-2h)-\- + cos(fl+n^ft)] ...(I) Now let S=^cos< 2 +cos Ca+A)+cos (a+2A)+ -i-cos (a+^l h). 140 A TEXT BOOK OF ELEMENTARY CALCULUS Multiplying both sides by 2 sin we get ^ h h 2 sin S=2sin cos a+2 sin cos (a+A)4-‘” ...+sin— cos(tf+n— 1 h) sin ( 2 ) ( '>+^)—sin ^ + or S= ;in r sin ( a+2n— 1 +sin^ a+2n— 1 y )~sm^ fl+2ii— ^ -)-sm ( a— ^ ) 2 sin This gives (1) as r COS X dx— Lt /I^O sin ^ a+2n — I ~ sin ^ '*~y ) sin h ih / = Lt //-►O sin ^a+«A — -y sin ^ a- — (Please note this step) h 2 sin =sin (a+6 — a)“Sin a sin Lt h 2 — 1 and /l-►0 =sin 6— sin a. Note. Integration as the limit of a sum is also known as the integration or the integration ab-initio. Exercise XXIII Integrate the following as limits of sums ; • J: X dx. 2 . J! m* dx. I. X* dx. (K.U. J977> INTEGRATION 141 4. 7. sin X dx. dx. 9. / jc dx. 10. j* e^dx. [ ‘i: cos’* X dx. cos X dx. '1 b a Hint. In the answer put a=0, [Hint. In the answer put a- sin* x dx. 0, b=x] Statement of Maclaurin’s Theorem 91. Statement. Lei fix) be a function ofjc which can be expanded in ascending powers of x. Let us assume further that the expansion is differentiable term by terra any number of times then * /(x)=/(0)+x/'(0)+^/'(0) + ^/"(0) + .Y + 7pr-/"(0)+... Proof.* Suppose f (a- ) = Oo + A- + a, -V” + + Then by successive differentiation, we get / X'X) = Ui + la-iX + 3fl3.Y^ + 4aip ^ + fix) = 2aa -h 2. 3.a3A'+ 3.4.04.x^+ /"(A)^=2.3.fl3+2.3.4.a4A-|- Putting A-=0 in (I), (2), (3), (4), we get /(0) = flo, /'(0) = *3„ fiO) = a., (2 !J, /"(0) = a3 (3 !) and so on. Substituting the values of 00 , 01 , 0^,03 in the R.H.S. of (1), we get /(A)=/(0) + A/H0)n-^ + /"(0)+ 9'2. Application to Simple Expansion The method has been explained with the help of the following worked out examples ; Example 1. Expand (0 e*, and (11) log (I + .t) with the help of Maclaurin s Theorem. *Tliis proof IS by no means a rigorous one. No proof for ihis theorem has, however, been prescribed in the Univcrsiiy syllabus. STATEMENT OF MACLAURIN’S THEOREM 143 Sol. (/) Let/(x) Z'U) fix) rix) — e tlien —e =e^ e fiO) f’iO) /'(O) /"(O) = e‘'= = e»= = c®= =^«= and so on Hence /(x)=/( 0 )+x/'( 0 )+^/'( 0 )+-^/"( 0 )+ gives e*=l + A'.i + yj- '^"^31 ^ *4 = l + x+yy+Tr .1 + t+|-) = COS -::^ = gives fM=f(o)+xf 'm+-^f(o)+-^ /''(o)+ cos x=l+^( 0 ) + .i^- (_i)+^( 0 ) + ^(l) +Tr+ =1-— + 4 ! Example 3 Apply Maclaurin’s Theorem to expand tan jc as far as the term in at. {KU 1977) Sol. Here /(x)= tan X /(0)=0 /'(.T) = sec*x /'(0) = 1 /'(■*) = 2 sec- X tan X /'(0)=0 f"'(x)=4 sec- X tan“ x+2 sec* x /"( 0)=2 /"(x)= 16 sec* X tan xH-8 sec^ x tan* x /"( 0)=0 /'"(x)=16 sec* x+88 sec* x tan* x+ 16 sec* x tan* x /'"(0)=16 and so on. Hence /(x)=/(0) + x/ '(0)+ -^/ '(.v)4- gives tan.v=0+x(l) + ^(0)+ (2) 2 + 15 ' + + ^ (0) + ^(16)+ 4. Obtain the expansion ^ Sol. Here /(0)=e«-l (K.U. 1976) -••(I) STATEMENT OF MACLAXmiN’S THEOREM 145 Let denote / (x) by so that /(0)=(y)o /'(0)=( yxV /"(0)={ y.)o. /"(0)=(y3)o and so on. Now, (1) gives y—Q'i= a A sin~i X ay V l-x* = ...(2) V\-x- (Please note this step) or get have / X ( y)o <2-1 (>’i)o= — I Also (2) gives (I— X*) V — Differentiating again and removing the common factor 2>’i, we (1— ...(3) Putting x=0 in (3), we gel ( y2)a=a- ( y)o=a“ • l = a\ Again, differentiating (3) n limes by Leibnitz's Theorem, we (1-x^) ynf>-(2rt+l)x>'„t, -(n^+a^) y„=0 [See Ex. [2) after Art. 5'4] Putting x=0, we get ( .V«+2)o=(n*+fi^)( yJo Hence, putting n = l, 2, 3, we get ( >’3)o=fl(I“+a’*) ( yl)^=a^l-^-a^) and so on. Hence gives /Uj=/(0) + x/'(0M-yj- /''(0)+ e-i iin-- a: Z 2! ■■ ■ 3 ! 146 A TEXT BOOK OF ELEMENTARY CALCULUS 1 . 2. 3. 4. 5. 6 . 7. 8 . 9 . 10 . 11 . 12 . 13. 14. Apply Maclaurin’s Theorem to show that ; ^ ^ "'^2! 3« +4l — a"*=l + (m log a) x+ 2 ! •.v®+ /.X . a:® , X* (i) sin x=x—-:ri--i- 3 ! 07 ) cosa:= 1 -^ + -^ 2 5 ! 07 /) tan x=x tan”^ x=x~ .sin X A-^ - + x^ x 15 •;c^+ 3 +-r- — l+X^ — .X* 8 + -e log sec .V=iA* + -r^ + -— + 12 ^ 45 .3 log(H-e')=log2+iY+l . .v 2 ! 5 x* 4 * sec.v=l + — 0 ) sinh x=x-\-~ + -^-{- (i 7 ) cosh A=l + -^+-^ + Hint. sinh a= e--e etc. 7 log (1 + tan .x)=A — ^ + §A*— log (1 + sin a) — A— -^-j- •' 6 12 ^ sin-> A=A + i + i . 3 .^- 4 - log ( 14 -a 4 -a-^) = a+ 2 a^ .v< + log (H-a4 A^>=Iog etc. 1 STATEMENT OF MACLAURIN’S THEOREM 147 15. 16. cos X ^ ‘ 2 ! 3 ! 2 ! 3 ! 17. cos b.\= 1 +ax+(a^-d~) ~ i ■. , ^ 2! ^ Tt ^ + 18. sin On sin"^ y)— y?r > , 3 ! 5~j •X''+ 19. (l+x)l+^=l + .y + ;c2-|-^_^ 20 . sin fe^-l)=x-h^-^x*+ 2 24 21. log (1-log (1-a:)}=;c+-^_ 22. e’" ^^-2) 3 I 23. e* _L ^ . e" + l 9 3. Evaluation of some limits with the help of expansion r, • is explained below by means of some solved examnlo^ It IS not, however, always practicable to apply the method. Example 1. Evaluate : sin 0 H 0-40 ^ Sol. 1 * sin 0 Lt -g-= Lt 0-40 *’ 0-40 0 — c® 0^ + T1 — 31^5! e (Expanding sin 0) = Lt 0 - 4 O Example 2. Evaluate : /, 02 04 \ 31+5! “ ) = !■ Lt x^O ^ Sol. Lt Lt I + X (log o) 4--^ (log a)2+ . . . . _ 1 ;r^0 ^ ^-►0 = i;‘„[ '»S‘>+^(loga)»+ 148 A ItXT BOOK OF ELEMENTARY CALCULUS ie 3. Evaluate : Lt — j:-4C •* Sol. Lt = Lt x ->0 ^ x -^0 - )-'■ Example 4. Evaluate : Lt log 4. Lt (c"* cos x). ,2 cos^ X t ^ ♦ s. Lt -..4-n“ 1. * L yV J 6 Lt 1 1 4-sin X \ 7. Lt sin A* x->in/2 V ^ / X 8. Lt sin 3x— sin x 9. Lt 1 1 V. x-#0 X j: •>— ' » 10. Lt sin^ X 11. Lt log X x-»-0 1— cos X 1-x 12. Lt 13. Lt ( x4 -iy-^" X— j :-*■ • \ X / 14 . Lt Y'* — /!'• 15. Lt V l+A-'‘ — X. A U 16. Lt ^''x-1 - j 1 17. Lt (cosec X— cot .v). x-*l V X— 1 A-*0 18. Lt sinh X 19. Lt cosh X — 1 a :-»0 X A-»-0 x^ 20. Lt tanh X x->0 X Exercise 2 Differentiate the following w.r.l x from first principles : 1. V sin X- 4. tan (x^). 7. f-f3^ 10. cos"^ (x*). 2. cos“ X. 5. log(Vx). Q ax+b CX + i/ 11. coscc*’ (ax). 3. y/\ogx. 6 . . 9. tan’Mv'^). ISO A TEXT BOOK OF ELEMENTARY CALCULUS 12. cot** (flx4-6). 13. log(ax-|-^). 14. cot (ox 4- ft) 15. X sin X. 16. X* tan X. 17. x” log X. 18. cos X cos 2x. 19. sin 3x sin x. ■ 20. X 1 ■ » • I ■ » • sin X Exercise 3 Differentiate the following w.r.t. x : 1. X 11 12. 15 log(^) 3. logx(xH-l). 2. X tan'^ X— log cos x+b 4. cos -‘C- 5. tan ‘ v;-s- + 6 cos X ) ‘ - r+i. ’■ — (SS)' 9. sin-M2x*— 1). S. + 10. COS”* (3x— 4 a:®). 1 + sec - (=:tT) + 2 sin'^ X. to X 13. 14. log tan ( + sin” X cos’* X. 16. tan U + vx^ 17. (log x)®*" ^ -l-(sin x)*°® ■* , 18. Find from : ax 9.x»+2Ax>’^-^^>>»+2gx^-2/>'+c=0, 19. Find dx • dy x"*y*=a 20. Find ^ from : dx y / 1 — x*+ •f 1 — ^=a(x+y). dy 21. Find ^ from : dx X— \/l— x*+>'l. REVISION EXERCISES 151 22, Differentiate : 23. Prove that the product of the values of from x^-\-y-=a^ and x>>=A 2 ]s unity Exercise 4 t. prove Leibnitz’s theorem on nth derivative hence find the nth derivative of : (a) X cos bx. (b) x\ax+b)^, (c) X log (flx+fe). (d)x^a^. (e) '2— x>»i+mV=0 (1 “X^)y,*+2+ xy„+i(2n+ l)+(m*+n*)j'o=0. If >'=(sin-^ x)^ prove that i\~-x~)y2—xyt = 2 ( I -x*)>„+ 2 -xy„^{2n + 1 )-n^y„= 0. If >'=[x+ V 14-jc*]", prove that ( 1 +x*)>'2+x;^i — rt*;;=0 ( 1 + x*)^„+jj+ x^„+i(2n + 1) = 0. If y=sin (m 6) and x=sm 0, show that ( 1 — x^)y2 — xy I + m^y = 0 ( 1 — x*)y„+2— (2n + I)x>»„+i — (m^—n*)y^ = 0. y—ef^^ sin bx, show that ^'s — 2o>', + (<3* +**)>*= 0. If sin (m sin*' x), show that il—x*)y2—xyi+m*y=0. 162 A TEXT BOOK OF ELEMENTARY CALCULUS Exercise 5 Evaluate the following 1 . 3. 5. 7. 9. 11 . { x ^+2 . )x*+4 W 1 —sin 6 1 + si + efe' X log X log log X 1 p — tan ; j 1 -r tan ; 1 sin 0 dx. I rfe. dx. dx. 2+x v3+x“''- 13. f ^ / dx. Jx^— 7x®H-14x— 8 j sin 2x cos* x dx. 15. 17. f Xy-3 ].v*-2x-l-2 dx. 19. [ L dx. 21. ]f^dx. 23. J Sin .V 25. 27. I— ^ J V l+.v* dx. 29. j (a^+g^-h.v“) dx. 31. r x+i J.y®-5x+6 dx. 2 . dx. f _i J l + COS^ X I dx. I 8 . 10 . 12 . \/ l+sin X dx. 1 cos x+sin X dx. ivi- I —cos X 4-cos X 1 dx. dx. Vx4-14- Vx rx*4-x4-4 J 11?+^ 16. f sin X sin 2x sin 3x dx. 18. 20 . Jsi 1 dx. I (x4-l)v A'*+2x 22. / cosec* X sec* x dx. 24. f 2x cot (.Y*) dx. dx. 26. 28. fsin (e^) J e-' H dx. 14* sin X rcos X dx. 30 . I y/x—x^.dx. Exercise 6 1. Find the condition that the line px4-^y=l should touch 2. Show that the tangent at any point of the curve REVISION EXERCISES 15? intercepted between the axes is divided in a constant ratio at tbe point of contact. 3. Show that the area of the triangle formed by the tangent at any point of the curve xy=a- and the axes is constant. 4. Find the length of normal drawn from (0. 0) upon the tangent at point ‘9’ on the cu ve .x=usin^9 and cos® 9. 5. Find the angle between 4) I 1 and 4 - = 1 . a~ 6. Find the tangent at (4. 8) to the = and find where it cuts the curve again. Find tangeni and normJ to the curve at the other point. 7. Find the normal to making equal intercepts wlh the axes. 8. Show that the subrnormul to y- = 4ax is constant. 9. Find the length of normal at any point for the curve x=a cos sin ’t. , 10. Water in a conical cup is rising at the rale ot - second. Find the rale of increase of its volume when li (height) feel and * (semi-vertical angle) = 45 . ^ 11 The radius of a .sater npplc is increasing at the rale ol I per second. Find the rate of increase of its area when its radius = 2 feet. 12. A lank is being filled up by water so that the svater in the lank rises at the rale of 3" per second. Find the rate of increase ot its volume when the length and breadth of the tank are 3 ant respectively. 13. A particle moves subject to the law .v-‘ A cos (»w), where X and t are distance and time respectively. Show that the velocity becomes zero after Show that the acceleration is proportional w to the distance. 14. A particle moves under a law V = K\^ Show that its acceleration is proportional to the distance covered. •KASHMIR UNIVERSITY PAPERS 1976 {For N£w Course Candidates) Section A 1. ^(a) Define the limit of a function /U) when x-* a. Does ^ ■' exists. Give reasons. 4 -►O (b) Evaluate any three of the following : Lt .¥-►0 stn X X Lt (Hi) Lt .Y . r ->0 (,v) Lt . .Y-..0 X 2. (a) Differentiate from first prineiples : Vtan x. (b) Differentiate w.r.t. x any two : (0 log A / l + si ' 1-si sm X sin X (ii) log X. (Hi) (tanA:)*+.Y‘®" -’'. .. t?' S^\ A has two of its vertices at (-a, 0) and (a, 0) and the third (.Y, y) moving along the liDe;;=/M;c. If A be its area, /j show that —7— —ma. dx x' dy (b) If y=x , . find 5 ;^ • X 4. (a) Prove that ~ ^ = I touches the curve j»=6 e wheie the curve crosses the ^-axis. (b) Find the equations of tangent and normal at any point of the curve x=a (1 + sin r), y=^a(l cos /). Section B 5. (a) Find the /ith derivative of e^ log x. (d) If j. = (log {x+ V l+x‘)?, find the value of>>„ at x=0. KASHMIR UNIVERSITY PAPERS 155 f /■ -I’t Theorem and obtain the expan- •Sion of /(.Y)=e"'''- ^ (b) Integrate any fwo of the following ; " Irrfc - a sim ;c (in) je' L±£J log X 7. Integrate any rhree of the following : (/) / e^(x sin a:+cos x) dx. (//) / sin" .y dx iiii) f sin ^ ^ / J V a+ 1 . / a (iv) / .Y COf‘ .Y dx. n ;5 integral as the limit of a sum. Find the limits as n increases indefinitely of any two of the following : (/) n n®-hl -+ '' 2 «®+22 Hi) „2 ■ m -L + 1 + 1 n \/(n*-l®) V(n®- n + + 1 (For Old Course Candidares) 4. (a) Illustrate tlie idea of a limit by examples. (^>) Evaluate any three of the following : .y3 — 9 (/) Lt - . ^-3 (1/7) Lt a'-\ Ui) Lt + . x^Q X c \ I . ‘^ot 6 — COS 0 Lt 5-;- — 0->«/2 COS® U 5. id) Differentiate ab initio sin \/ x or cos- .y. (6) Find of any three ; (0 > = V (11) x= tan"* __ 1+/® 1 + /® — co s X + COS X ^at^ ("'■) y= x-i- x+ <'v) ^=Vl^x to oo 156 A TEXT BOOK OF ELEMENTARY CALCULUS (a) Differentiate (any three) : (0 tan-^ 2x l-x* tan X w.r.t. sin'^ 2x 1+x* (/i) (tan ^ w.r.t x. (Hi) x^ w.r.t x. (iv) X COS“‘ X V 1— A* W.r.t. X. ib) If siny=xsin (a+y), prove that ^ ax "••• - sm a Or (b) Prove that — + a ■^=2 touches the curve ("" ) f =2 at (a, 6) for all values of «. ^ 7. (a) If a point is moving in a straight line and distance in feet irom a point in the line after t seconds is given by Y=5H-2r+4l* find the acceleration at the end of 3J seconds. * (b) Find the equation of the normals to the curve (v— 3)(v— 6) =a:* at the point where ^ , 8. (a) Find the /ith derivative of cos® x or 3x+7 (b) If 3 ;=[.ic+ V" l + x®]"*, show that ; O') >'2(I + .^) + x>>,— (ii) y»+2(l+x®) + (2.ic+ I)xy„+|+(/i*+m*)>',.=0. 9. Integrate (any three) : a) x*+2a-+3 x+1 (///) a / £± 5 . V a—x (ii) Vl+sinx (iv) sin*‘ X. (v) x» X— 1977 Section G 5. (d) Explain with regard to a function f(x) the statement a and x-*^a giving a few examples. (b) Evaluate any three of the following : X" — a" X-^Q ^ 0-0 Lt ^^oVsin qx } (Hi) Lt (iv) Lt X -¥0 e*-l KASHMIR UNIVERSITY PAPERS 157 6. {a) Define differential co-efficient and differentiate from &st principle — • ^ x+c ib) Differentiate w.r.t. x any two of the following / 1 \» 1+—^ (0 (, + i) +(,) tan-» ^ \ \cos X— sin X ) (iv) log I (Hi) cos (x^). 7. (a) The diameter of an expanding smoke ring at time / is proportional to /*. If the diameter is 6 cm after 6 seconds, at what rate is it then changing ? (b) If V 1 — x'^+ V prove that dy _ x/ l->.2 jx V 8. (a) Give the geome*rtcal interpretation of the derivative of a function and use it to obtain the equations of tangent and normal to the curve y—fix) at a point (x, >’) on it. (b) Find the equations of the tangent and the normal at the point “r” to the curve whose equations are x=a cos® /, sin® t. Section D 9. (a) Find the nth derivative of e* cos* x, _L (b) If y"' +y =2x, prove that (x*— 1)>',.+2+(2 x+ l)xyn+i+(n®— /M®)y„= 0 . 10. (a) Apply Maclaurin’s Theorem to expand tan x as far as the terra in x*. (b) Integrate any two of the following : (i) i dx H'Sin X* « 1 ^ X* dx +x® (Hi) f - .. = — - (iv) J cos® X dx, J sin^ x+t^ cos* x A TEXT BOOK OF ELEMENTARY CALCULUS 11. Evaluate any three ; { xe dx (»0 I X sin -^ X dx (v) / cosec* X dx. (m) / X tan“* X dx. (IV) S e" cos (bx+c)dx. (d) Define as the limit of a sum and evaluate X* dx as the limit of the sum. (b) If y=sin (sin x), prove that tan x-\-y cos* x=0. ANSWERS Exercise II 2. 3, -6, -5. H-2x- x2 -5x2 a^+2ak+h~-\-2a-\-2h- -5. 3. 1. V3 2 . 0. 7* 2a4-/; 15. (0 x= 1, (|7) x= = 2, 3. (/77) Ax. 16. 2n, 2 Irr, n Exercise III 1. 10. 2 3- 3. 4. I. 5. a 6 • 6 3 4 • 7. L 8. 1. 9. 2. 10. 0. 11. 1 12. a l-y- 13. i 14. 15. m n 16. sec- ■ X 17. i- 18. it 19. 1 a • 20. 26, m n # 27. 0. 28. 1. 29. . 1 25 • 30. I. Exercise IV 1. U) 5x^. 00 1 x» ■ Uii) I 2v X Uv) 3x^ (v) 2 (V/) 20(2x-|- ( :vii) fi 3 (8- 13x)~* # (v/77) - -{(5x+7) 2. (/) l-2x. (m) 4 a“+15x^-14x+4 (v/) m. 3. (/) mpx^-^ - Uii) -2-16/ 12/2. 4. (i) 1 5x2. (/■/) 3x*-12x+ll. (/V) / -n 28 56 (v) 2x-f-3- _8 i X Hi) 6 (3x-f-4). 07) 4x=*+12x*+I2x+4. (/V-) 27. (Oil. 1 1 077) - V x^ .160 A TEXT BOOK OF ELEMENTARY CALCULUS (wi) (x) (xiii) 2oa+Z;. (v) 3 (3x + l)2 ' (/V) 1 3. -^T* 16 ^u/lf \ 1 Ox) (5X+7F 2\^ax+ft* 2vxV ^ a fxil (Xii) ■7v_u 18 2 . v/ (ax+ 6)®* (cx+rf)** ^ ^{x+3y 1 1 2 (x+c )»'2 Exercise V 1 . 3. 5. 8 ( 3 ,v 2 + 7 x-}- 5 )^( 6 x+ 7 ) y/2-x* -4(16x-7) 7x+9)*'* 2 . 4, niax^^bx+c)”~H 2 ax+b) - 6 ( 7 x«+ 12 xg+ 2 n (x’-i-4x*4-21x)» ' m~n 6 . -^( 3 px*— 29 x+r)(px^— " p-q 7. (2x= + 3)(x*+3x>-l) 9. _ 1 14x 5 ^'( 7 x— 3 )* 10. 20.v(c2+x2Xc“+(c*+jc’«)2]^ 11. 30 a^-27.x® + 8a:. 13. 5.v^-|-21a'3 + 20x. 14. + 1 -15. 2(6.\'^~a— 5). -1 17 8 ^ 3^2 _ 3 < X 10 •x( 36 a‘^+ 56 .v^fl-“ 2 c) (A=+a)3 1 6 — lac 2 ' (v — 2^7)»'H^— 21 . 18. -6a(3.v2+4)(1-a2)(5a*+2). 20 . \ 22 ~2(a-H- 2jc+5) 23. a\a- ' a^-x ^] x^Va^—x* 24. (A--h2x-3)* -x4 - V' A^— I ANSWERS 161 * / 2-1 4. -i. 2 . - Exercise VI b 1-/2 18. l-2r» ■ 7. - ^'8 8 . - ^/3 10. hx-\-by 11. ^ 12 13. OA+Ay+g 14. i X"-^. 15. ftA+6>'+/ • 16. l-A® 17. A+Vl-A® . 3a®(1+a®)* .X ad- be {c'x+d’)^ a'd‘+b'c‘ * (ax+6)2 Exercise VII 1 t JC+g J'+/ x»- /- 55x*-22.y 14a- 15 ,, l + WA^‘-«-^-(/i+l)x (l-xr- I. 1. —m sin mx. 2. j-^cosx'. 3. —m cos”"* A sin a. 4. 9 tan® (3 a+ 6) sec* (3a+6) 5. %a{ax~{-b) sin^ (ax-hbf cos (oa+6)*. 6. cos A sec* (sin a). 7. 2a sec® A tan a. 8. m sin’** * A cos"+* a— n ; sin«+i X cos"‘‘ A. 9. — 2 cos A (a^—b-) sin a (1 + sin A)® * (6 + flCOSAj® II. 1. — i cosec® 2. i. 3. 1. 4. 3 sec“ 3 a. S. J. 6. 1. 7. -J. 8. -i. in. 1 ^ ? 2 1+A® 1+X»* '••Vi 3 -A® 5. 1 ^ 1 2(1 +a2)’ 1+A n 2a ** 1+A* IV. 1. sec* A. 2. 2a® cos A* 3. —cot X. 5. 1. _ 6. \ a 2 cos a^ sin® A cos A 7 . J 8 . A XNs 162 A TEXT BOOK OP ELEMENTARY CALCULUS 1. i COS X 3. V sin x* sec* Vx tan X x b cos X 4\/ Mtv/sin -s/x 1. — ~ cot 9. 3- cos 9. 4 X sec* v/ 1 2 ^ (I+A-*) tan V 1 + ^ 2. tan 0. o 3a 4. tan 0 IX. 1. 3. 3 sin 3.V. 1 2va cos vx. sin .Y 5. 2 V cos X 7. sin x-\-x cos x 2. 5 sec 5x tan 5x. — 2.Y cos X sin x <5. 2x sec* ,v*. 8 . sec* .V 2 v'tan A- VIII I. 1 A. V 1-9a* ■ 2. 3 I 4. A*+o*‘ 5. 1. 6. V I — (3a- + 2)* 2 a: Vat^— 1 sec* (tan“* x) l + x* 7. _ 1 2a- Va»-1 8 . 0 . 9. sec* A ian“* a+ 1 +A* 10 . a-\-b cos X U 1 4.v*-i-12A+1 0 2a * 1 IV. 1. _ A- V 1 -hA*' 2 1 3. I 3. * . a-\-b cos A 2 V1-a“ ^ COS-^ A — A VI A* (1 ~A*)*/* ANSWERS Exercise IX I 1. 2.Y \ 1~A- 4. sec .Y. 3. sec .Y. 5. 2 cosec X. 6. — +sec x cosec 7. 9. I (I— x'-*) ian“^ X x^-l 8 1 x*-4 n. 1. {x-\-\)e\ 3. t* (tan x+sec^ x). 5. e* cos e . 6. 2 sec^ (2x+3)o‘®” (2(+3) ^ X log X 10 . cosec X. 2. ^/e“^+2e-'. 4. — (e^— sin ,sin'* X Vl-x' 5- log fl. 8. ^tan"^ x+ l+x -.) ,x tan”i X. 9. IV. 1 . V 1+x* ,V li-x- . 13. 3x^+e^ 1 3. xll+dog x)2j • cos (log ^ ^ ) 2 QOS (log tan a) sin X cos .Y x(x+l) 1-x* • l+x-'+x^* 6 . lah a* cos' X— 6* sin* x* 5. 7. -12 x3+3 x2-3 ( >_3ja X («)* ^x(logo) 1 X log X log (log x) Ex^rcisc X (Section A) 1 . 2. 3. 4 sin xXlog xXe*x \/xx j cot xH ^ — ^_ 1 _ 2 £±_L? ( X log X ^ 2x ) i \/x V'sin X Vlog xi— +cot xH ^ i X X log X ) X* (l+x+x log x). (tan x)*^* ^ . cosec* x (1 — log tan x). X. A TEXT BOOK OF ELEMEl^TARY CALCULUS 164 (sin"^ xY \ log sin * x-\ C sin"* ; X X r+ si sin x+cos X log x |* (sin sin .v+cot x logxj- —x^ log ex . sin A* fl**® log a . log b. x" X^ |^ + logA+(log x)“J- II. .X- (I+Iog -v) + (^ j .Y* ' (I-log a:). 7. 8 . 9. 10 . 12. 13. 14 15 16. (sin (cot X cos x — sin x log .x)+(cos ■* X (—tan x sin ac+cos x log cos (sec ^ (sec a'— cosec x cot x log sec x) + (cosec x)^^' •' (sec x tan x log cosec x— cosec (tan x) (log tan x+2x cosec Xx) _|_xtan ^ jQg x+-^tan xj* 2' cot X r. ... — (log 2 — 2 cosec* X -i] X cos V f± 1 X 1 1 x^ L ^ V 1 — X* cos * X 1 — •X*J Section (B) 1 . 2 . 3. 4. cosh* X ~~-^l -i-cos X cosh x) isinh x+sin x)* tanh X, ^-r — ^ sinh .Y cosh x sech X for each part. ^sinh A j;o5]j ^ [Qg 5. sec X for each part Exercise XI 1. 3024.v\ 3. — 120fl^ (ax+6)“ 2 . - 4 _ X 13 -.Y) 8 ANSWERS 165 5 . 6 . 14 . 4 tan^ X sec“ x+2 sec* x— 4 cot- x coscc^ x— 2 cosec* x. K , — — (x“— 1 (xS— l)3i X -Ij ■ _a X • 15, cosec^ 0, Exercise XII 1 (- 1 )" IJL * «! ‘ (I+xr+> • 3 _ 6" . B ! (-ir ^(b !) 3" -^ (fl-6x)'‘--*‘ • (3.T+7r-^* 5 (-!)>■ |_^4-i . 2" (2« + 3r'* 6. i cosCx+iwit)-] . 3" . (3x+iB-). 7. (6-a) . 2"-ixcos (2x+i/m). 8. 2'' +5" (log 5r. H2'‘cos (2x+A/iTt) + 4'' cos (4x+i/m) + 6' cos (6x4'*Mrt)}. 10. -2'*-! cos (2n+4B,c). 11- ^{4" cos (4x+inrt)-f 2"+* cos (2x+^m)}. 12 . Je"{2’‘ '2 cos (x^-i/I:t)-|-lO"/^ eos Ox+b tan-’ 3)}. 13. .'.2"- [cos (.+ ^) + „n(,v+-^^)], 14. . COS (x sin a +na). s 15. 3 sin (x+n sin (3.v+-^) 16. ie"+iY"e"cos (2x+Ht)} wliere Y = VS and 0-=tan-i 2. 17. h" sin (x+«0)~iR'' e" sin ( 3x-\-n^) where Y=v^2, 0=2L 4 R= 10, ± ) J- 20. 5" sin (4 x4-b Ian*' ^). 166 A TEXT BOOK OF ELEMENTARY CALCULUS xin I. ^ 3H(n— DJ cos , “!-«f3A-~ (n— — 2)] sin 2. 2(-ir-i . (n-3) ! 3. ^ [log x+^Cj "C, A- =+ + (-!)'■ 1 . (h — 1) ! .Y'"]. 4. a’' [x^ (log ay‘ + 2nx (log a)"-* + /i(/J— l)(log 0 )"-=^]: 5. e-' [x=a'' + 2/7x Dn"-*]. A-"- = -f 7. (a=+/, 2 )'/-i , sin |^Z)A + m tan * ^j+nCa^H- ^ xe-'sin |6.Y+(n-l) tan-* -|-|- 11. When fi is odd, ( j, )0 is !(n-2)=+a”){(n-4)>+a=}...{(3»+a=)(I + o=) . a}. When n is even, ( Vri)0 is {(/f-2)2+a2){{n-4)^ + o=} {4a+aa)(22+aS) 0 =. 14. When /i is odd, ( 3',.)0=(//i-— (n — 2)=}{/na— (n — 4)*} (m*— 32)X (m’*— 1) When n is even, ( >-..)0 = (rn ^ — (rt - 2)n{m« — (n - 4)=>} (m^ — 2>)nj=. Exercise XIV 1. (0 2tiy. (ii) 2n(Y+/i), (ni) 4:ri'*. 2, — ake'^* sin cos (ti»/+< 3 ). *> '3. 4. 72Tr square inches per second. 5, ( 1 ) 92J ft./sec. (//) 84 ft. 'sec*. (m) 150 ft 11. i/. Exercise XV 1. (i) >'Y=2a(X+x) ; 2a(Y- v)+r(X-.Y)=0. (i7) a*Y - 3 a-*X + 2a^= 0 ; 3A*(Y->-)-fa“(X-.Y)=0. (Hi) (.Y*— av)X + ( }^—ax)y =axy ; (Y—y)(x-—ay)~(X~x)( ya—ax)=0. (iv) ty=x-{-at* ; y-\-ix=2at~i~ai^. (v) bx—ay sin ^—ab cos 6 ; ax sin 6+6>*=(a*-{-^2) tan 0. ANSWERS 167 4. 8 . !3. 1 (v/) sin ^t—y cos ^t=at sin ; x cos it-hy sin i/=or cos i/+2a sin i/. 2. U-) ^+J2:=i . / (Ar-X'j= ^ Tangent at (a cos 0, b sin 6) is X cos 0 , sin 0 a 6 “‘• Normal at (a cos 9, b sin 6) is ax sec 0— 6^^ cosec 0=a*— ft*. (/V) Tangent ; jc>+y.x=2c2 Norma) : xx' —yy'—x'^—y'^, iiii) Tangent : 3x+2>>=3 Normal; 2x+3>*— 15=0, (iv) Tangent ; ^”jc+a=0 Normal : ;c+>»=3a. (v) Tangent ; x=2a Normal : y=0. (2. 16), (2, -16). x-^^y=±9. 7. y = 2x-\-3. tl. a«/3+62m2=/i». m .*n-l (a cos a)"*-»+f6 sin «)«-»=y Exercise XVI n T* 3. tan-> 3. 5. tan -(i 9. 90* ; tan y :tan->(-|-) tan-* (^ 16 ). 8. tan-* (if). ) 3 22/3 Exercise XVII a sin* 0, a tan 0 sin* 6, a sin* 0 cos 6, a sin* 0 tan 0. Exercise XVIII 1 o^i/i . (^8)* . (2 a;+3)« g , X- , iSX , , — — 2. (/) - 1 12(3x+4)*- 12 «) f- 4 +i.-. » — 168 A TEXT BOOK OP ELEMENTARY CALCULUS (i/‘0 3x+2x®H-5 log X. I 1,1,3 ^ X 2x»+3x»'^4x* (v) — +Iog X. 2cx* cx ® 1 (viV) log (a— 6x). (lv) ^ ^ + f X* + 3 log X. (x) I (x+ !)«'*+ 2{x+l)*« (^0 {(x+a)“'»-(j:+6)*'>}. (vi) log (x— 5). (viii) 0-8«+e» — ^+3 (x/l) — 1 3a (ox+6)® 3. (i) x4-log (x— 4)®. (ill) X— I log (3x-l-4). x» 00 4x-|-log (x— 4)^^. (iv) - |x»+9x-34 log (x+3): (v) Jx“— fx-l-l log (2x+l) .2 (vi) -~4-x-l-2 1og(x-hl). (vii) 2 x® + Jx*+9x. (viii) |(.v+3)=*'»-6(x+3)>« (ix) §(:^+a)»^2+2(6-a)(x+a)i« (x) l(x+3)»'^-2(x+3)®/*. 4. (i) 3 sin . (I'ii) cos X. . V tan 4x (v) — (vii) tan x— x. (ix) xH =— (xi) i tan 0. (xiii) — i ^^^^^+co8 x^. (ii) cos X. (iv) tan X. (iv) tan ax I II 4 a (viii) iV (sin 3x+9 sin x)- (x) tan X— 'Cot x. (xii) tanx+cotx» sin 5x (x/v) ~ “|“Sin X ) ANSWERS (JCV) K COS ( p+q) X . cos(/)— o)x p-\-q + ■] ixvi) cosec X— cot x. (xvm) X— tan x+sec x. (xx) 2v/2sin p-q (xv/i) —(cot x+cosee x). X (x/x) 2 ^sin y — cos 5. (0 -e-». (iv) (») (v) —a (m .2* +2x. (vi) .n+1 + n «+l log n -t-x. 6. (/•) i. (/) h iiii) (/v) J log 3. (v) i log 5. (Vi) -y • (v/7) V3. (Vf7/) -r- V3 Exercise XIX (i) J tan'* y. (/;) i tan-* (i7/) i tan-* (2x). (iv) T^tan-*;^^* 4o 4o (v) tan"* (x+2). (V,) 2^2 tan- 2^2 (i) sin"* X. (/7) sin-* y- (///) v^2sin-*^ (iv) — sin-* (ax), a iv) sec-* V3 (vi) sin’^ x(x^i). (V/O sin-(*+2_y . (/) sec-* X. (/7) sec-* (3x). (///) i sec-* (2x). (iv) 2 sec'* -y- 10^ log 10 170 A TEXT BOOK OP ELEMENTARY CALCULUS 4 (;■) — COS“* X— v/l — X®. (//) V a*— X®— cos ■(f> (Hi) fTt-2) a* 4 5. (/) log (x*-|-4). (it) i log (3x2+ 2x -I). (m) log (tan-* x). (iv) log tan x. (v) log (log x). (vi) log (e^—€~^). (vii) log (x + cos x). (viVi) —log cot X. iix) ^ log (a-\-b cos x). (x) log (sin * x). (x/) sin (tan * x). ixii) tan-* (sin x). (xiii) log 2. (xiv) log (I + Iog x). (xv) — i log (3 cos* x+2 6. (i) tan * X e (ii) c'. (tan'* x)® 1 1 Uii) 3 l-«^log V* ' » (v) (log x)"-* (vi) (V//) tan'* X®. , ...v tan* X (vi/i) — (IX) log (log sin x). (x) tan * (sin x). (Xl) —cos (log x). .... cos* X (xii) — 5 cos X. ixiit) tan 9 e (xiv) sin X— 1 sin* x+i sin (XV) 2 sin v'x. (.w/) . 4 ixvii) ixviu) - COS^ X I COS^ X (xix) — cos* X ANSWERS 3 . 4 . 5. 6 . 7. 10 . i log tan [-i-( .v+tan- )+ sin a . log sin (.v— a) + (jc— cos a. (0 logi±_^±±Z. Hi) log a \ tan"* (2 tan x). 1 I. \/2 tan B tan ( tan .t \ V 2 ) V 6- 12 J tan - (iS^). n. n 2V5 12. -i 1 13 14. 15 . 2 tan .Y + 1 (l 4 -v 2 ) 3/2 2 -f (H';c=P’*+(l +;cs)*';\ +f sin- a ^ log 5i" sin (a— d) ®sin(x— 16. n. 2 ( sm - cos y ) - V 2 log tan (y -Hy j 18 . 19. 20 . L^log (a-[-b cos x) — rr , — r* o® **(< 2+6 cos x) Klog (sec x+tan x)]®. i log (■^) *og (ab). Exercise XXI I- i log x -2 x +2 <> 1 1 2^-1 • n" ®®3x+4 3. -3 log (x+1)-— L +4 log (x-2). i tan"*y +i 8 tan“* X 2 • 5. log (x+l)(x-l) i log 1 +x* 1 -x* • 7. log (x— l) + i log (x“+l)+tan- 8 . 9. 4 log (x-I)-i log {x+3)- A» J iog(*+i)-iogu+3) l+y ij^rr 172 10. 13. A TEXT BOOK OF ELEMENTARY CALCULUS V a® (^b)(a-c) 11. Iog(l+e-*)-c~*. i log (l-fcos x) — i log (1-cos jr) — f log (1+cos 2x) (.v3-3A:2+6Ar-5)e". 2. f logx-^ X cos .r+sin X. 4. x tan j:— log sec x. X sec X— log (sec jc+tan x). 6. ;cIog x— ;c. X® . , ysin ^AT+i- VI— a:®— > (1— ;t*)3/2 1 (.v^-1) tan-> x- A (x^-3x). B (sin 2.V— 2x cos 2x). 10. —cos x log cos jc+cos x, X (log x)^—2x log .r+2.r. AT sin j ^2 13. Af cosh X— sinh AT. 'l2_4!±£l-^ log-l±2f_^±S^. IS. x+ V 1<». -^sin (4Ar-tan-> J). 17. e* tan-i log (1+e*') 19. _COl3 X +cot .x+1. 1 a tan * x — e a 20. X — Vl— x®sin“'x. 21. c* cot * cosec X cot x+ J log tan tI X cos v/2 ( M-g- 2 3 I , sin 2g— sin 2b ~(b-a) + 5 e~ sin X. Exercise XXIll -% m“— /w^ A. log m 4. cos o— cos ^ b — a sin 2g— sin 2b 6 . ^ 5. 6 . .-I =. ANSWERS 173 7 . «• 1 - 9 - i Revision Exercise 1 1. log a. 2. 1. 3. i. 4. 0. 5. 9. i. 0. 6. 10. 0. 0. 7. 11. 0. -1. 8, 12. 2. 13. e. 14. n 15. 0. 16. s. 17. 0. 18. L 19. 20. 1. Revision Exercise 2 1. sin X 2. — 3 cos* X sin x. 3. 1 4. 2x sec* (x^). 2xv log X 5. 1 2x* 6. 2xe-''“- 7. X 8. ad— be vi+x^ (cx+i/)* * 9. 1 10. -2x 2vx(H-x)* 11. 1 12. —a ^Vfl2x*-1 l+(ax+6)* 13. a ax-{-b 14. —a cosec* {ax+b). 15. X cos x+sin X. 16. X* sec* xH-2x tan x. 17. x"-' (1-J-rt log x). 18. —M3 sin 3x+sin x] 19. 2 sin 4x--sin 2x. 20. sin X— X cos X sin* X Revision Exercise 3 1. 3. log X. X log X— (x+1) log (x+1) x(x+l)(log X)* 2. tan~* X, 174 A TEXT BOOK OF ELEMENTARY CALCULUS' 8 . 9. 12 . 6 . - a^b cos X 1 5. - 1+x* V 2 7. 2v^l— X* 3x* 1 1+x* l+x* >°8 Vl-x®* X 10 . - X -HI— log X X* 13. Vl-x* 1 11 . 0 . x(l— ylogx) " ' 2>»— r 15. sin”* X cos" x[m cot x—n tan x]. sin X 14. sec X. 16. -1 2, x(H-x) 17. (log x) Sin X [ X log X + cos X log log X ~|-H(sinx)’®‘*'*^ / cot X log X -H ^ J.-- (a.v + /i>-Hg) ''(hx+by+fr 19. ^ nx yZ. ^ 21 '' ' 22. -■*’ . iO. . , V V 1 — X® Vl-x* --S Revision Exercise 4 1. (a) c'’‘'[xr" cos Ci>x-Hn0)4 nr"** cos (!>x-H«fl — !>)] where r— v 9=tan"*'*/®^ (6) |n [.v-a'*4-2nx(ax-H6)a** *+ ^^”2 (nx-H6)® - xo"(— !)"■* In— 1 , no^'H— l)"**!i*— 2 (c) ■ + ( 0 X-H 6 ) (nx-H6)"-* (b) 5(-n®-* I « . 3"-* (3.Y-H7)"** £ 4 — FlO"/* cos (,3x-Hn tun * 3)'H3.2’'''® cos ^ (d) ix5"/* e’ sin C2x4-n tan * 2). ANSWERS 175 1 2 8 . 10 . 12 . 13. 14. 16. 17. 18 20 . 21 . 22 . 24 i[6" cos ( 6x+'f) + r cos ( 4;c+^ ] + 2" cos if) -|-[4.2"cos ( ='+'? )] cos { 2x+”^ ) + 4'‘ cos (-0" 14,-^.-,-^ 2)-+i + 2( (h) (-1) [ (4_ 3) '■ t J - (— J ■ (/) e* * sin (x sin Revision Exercise 5 |[tan-H.Y+l) + tan-Hx— 1)]. 1 _! 1 V2 tan cm a 2^ 2(aey log II sin 0). -( ^ log 7. log log log .V. 9. tog (cos .Y + bin ,v». U. 5(34-x)=‘/^-2^/^. 2 ' 2 log £/" log (4/e)' - ( 2 2 )■ ^2'°® '“"(f + t)- log (cosec* jr— cosec x cot x). !(x+l)*'*-Jx*'*. i log (x-D- J log (x-2) + i log (x-4). logx + itan * 15 . 1 r cos 6x cos 4x cos 2x“l 4 L 6 4 -'T'J* log (x=*-2x+2)-tan-» (x+1). sm-'(2x-I). 19 sin->(x-2). sec’ (x+1). f x+ 10 log (x— 1). — R cos* X. X* X* .v2 I I 4 ^ “3"+ 2 tan x-cotx. 23. x + sin 2x, log sin (x’). 25. Hlogx)*. 26. sin (e^). 27. V 1 + X* 28. Ian - - + 2 log see - - 2 176 A TEXT BOOK OF ELEMENTARY CALCULUS 29. 30. 31. a* a-hl J[sin-» (2x-l)-i-(2x-l)^/l~(2x-m. 4 log (jc— 3)— 3 log x(x—2). Revision Exercise 6 1. /w"*.«'‘==(m+n)*"+". p*". g". a”+". 4 . a sin 6 cos 0. 5. 90®. 6. (1, 1) ; 2>— 3x+l=*0 ; 3y+2jc-5=0. 7. A-|-3a = 0 : y+x— 3o=0. b 9. — y/a^ sin* G+A* cos* 0. 1®* y- cubic ft./sec. a.1. «/3 sq. ft./sec. 12. f cubic ft./sec.