Chapter 3 Differentiation PDF

60 128 Differentiation Introduction. E3 The rate of change of one quantity with respect to some another quantity has a great importance. For example, the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of velocity is called its accelera...

60 128 Differentiation Introduction. E3 The rate of change of one quantity with respect to some another quantity has a great importance. For example, the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of velocity is called its acceleration. The rate of change of a quantity ‘y’ with respect to another quantity ‘x’ is called the ID derivative or differential coefficient of y with respect to x. 3.1 Derivative at a Point. The derivative of a function at a point x  a is defined by f (a)  lim U limit exists and is finite) h 0 f (a  h)  f (a) (provided the h The above definition of derivative is also called derivative by first principle. D YG (1) Geometrical meaning of derivatives at a point: Consider the curve y  f (x ). Let f (x ) be differentiable at x  c. Let P(c, f (c)) be a point on the curve and Q (x , f (x )) be a neighbouring point on the curve. Then, Slope of the chord PQ  f (x )  f (c). Taking limit as Q P , i.e., x c, x c we get lim (slope of the chord PQ)  lim x c Q[x, f(x)] ……(i) U Q P f (x )  f (c) x c y [c, f(c)]P As Q P , chord PQ becomes tangent at P. ST Therefore from (i), we have  f (x )  f (c)  df (x )  Slope of the tangent at P = lim.   x c x c  dx  x  c Note 0  f(x) – f(c) x– c x :  Thus, the derivatives of a function at a point x  c is the slope of the tangent to curve, y  f (x ) at point (c, f (c)). (2) Physical interpretation at a point : Let a particle moves in a straight line OX starting from O towards X. Clearly, the position of the particle at any instant would depend upon the time elapsed. In other words, the distance of the particle from O will be some function f of time t. P O t = t0 Q t = t0 + h X Differentiation 129 Let at any time t  t 0 , the particle be at P and after a further time h, it is at Q so that OP = f (t 0 ) and OQ  f (t 0  h). Hence, the average speed of the particle during the journey from P to Q is 60 f (t0  h)  f (t0 ) PQ , i.e.,  f (t0 , h). Taking the limit of f (t 0 , h) as h 0 , we get its instantaneous h h f (t  h)  f (t0 ) speed to be lim 0 , which is simply f (t 0 ). Thus, if f (t) gives the distance of a moving h0 h particle at time t, then the derivative of f at t  t 0 represents the instantaneous speed of the E3 particle at the point P, i.e., at time t  t 0. Important Tips  dy dx is d dx (y ) in which d dx is simply a symbol of operation and not ‘d’ divided by dx. ID  If f ( x 0 )  , the function is said to have an infinite derivative at the point x 0. In this case the line tangent to the curve of y = f(x) at the point x0 is perpendicular to the x-axis If f (2)  4 , f (2)  1, then lim x 2 (a) 1 (b) 2 Given f (2)  4, f (2)  1 (c) 3 D YG Solution: (b) xf (2)  2 f (x )  x 2 U Example: 1  lim x 2 x 2 f (x )  f (2) = f (2)  2 f (2)  4  2(1)  4  2  2 x 2 Trick : Applying L-Hospital rule, we get lim U x 2 f (2)  2 f (2)  2. 1 If f (x  y )  f (x ). f (y ) for all x and y and f (5)  2, f (0)  3, then f (5 ) will be [IIT 1981; Karnataka CET 2000; UPSEAT 2002; MP PET 2002; AIEEE 2002] ST (a) 2 Solution: (c) (d) – 2 xf (2)  2 f (x ) (x  2) f (2) xf (2)  2 f (2)  2 f (2)  2 f (x ) 2 f (x )  2 f (2)  lim  lim  lim x 2 x 2 x 2 x 2 x 2 x 2 x 2 = f (2)  2 lim Example: 2 [Rajasthan PET 1995, 2000] (b) 4 (c) 6 (d) 8 Let x  5, y  0  f (5  0)  f (5). f (0)  f (5)  f (5) f (0)  f (0)  1 Therefore, f (5)  lim h 0 f (5 ) f (h)  f (5) f (5  h)  f (5 )  f (h)  1   lim 2  = lim  h 0 h 0 h h  h  { f (5)  2 }  f (h)  f (0)  = 2 lim    2  f (0)  2  3  6. h 0  h  Example: 3 If f (a)  3, f (a)  2, g(a)  1, g(a)  4, then lim x a (a) – 5 (b) 10 g(x ) f (a)  g(a) f (x ) = x a (c) – 10 [MP PET 1997] (d) 5 130 Differentiation lim x a g(x ) f (a)  g(a) f (x ). We add and subtract g(a) f (a) in numerator x a = lim x a g(x ) f (a)  g(a) f (a)  g(a) f (a)  g(a) f (x )  f (x )  f (a)   g(x )  g(a)  = lim f (a)   lim g(a) x  a  x a x a    x  a  x a  f (x )  f (a)   g(x )  g(a)   g(a) lim  = f (a) lim  = f (a)g ' (a)  g(a) f ' (a) x a  x a  x  a  x  a  [by using first principle formula] 60 Solution: (b) = 3.4 – (–1)(–2) = 12 – 2 = 10 x a g(x ) f (a)  g(a) f (x ) x a Using L–Hospital’s rule, Limit = lim x a g ' (x ) f (a)  g(a) f ' (x ) ; 1 E3 Trick : lim Limit = g' (a) f (a)  g(a) f ' (a) = (4 )(3)  (1)(2) = 12 – 2 = 10. 1  dy  If 5 f (x )  3 f    x  2 and y  xf (x ) then  is equal to  x  dx  x 1 (a) 14 7 8 (c) 1 1  5 f (x )  3 f    x  2 x......(i) 1 1 1 in (i), 5 f    3 f (x )   2 x x x  ......(ii) D YG Replacing x by On solving equation (i) and (ii), we get, 16 f (x )  5 x   y  xf (x )  at x  1, (d) None of these U Solution: (b) (b) ID Example: 4 3 3  4 ,  16 f (x )  5  2 x x dy 1 3 1 3 (5 x   4 )  x. (5  2 )  f (x )  x f (x ) = 16 x 16 dx x dy 1 7 1  (5  3  4 )  (5  3) =. dx 16 8 16 3.2 Some Standard Differentiation. d n x  nx n 1 , x  R, n  R, x  0 dx ST (i) U (1) Differentiation of algebraic functions (ii) d 1 ( x)  dx 2 x (iii) d  1  n  n    n 1 dx  x  x (2) Differentiation of trigonometric functions : The following formulae can be applied directly while differentiating trigonometric functions (i) (iv) d sin x  cos x dx (ii) d sec x  sec x tan x dx d cos x   sin x dx (v) (iii) d tan x  sec 2 x dx d d cosec x  cosec x cot x (vi) cot x  cosec 2 x dx dx (3) Differentiation of logarithmic and exponential functions : The following formulae can be applied directly when differentiating logarithmic and exponential functions (i) d 1 log x  , for x > 0 dx x (ii) d x e  ex dx Differentiation 131 d x a  a x log a , for a > 0 dx (iii) (iv) d 1 , for x > 0, a> 0, a  1 log a x  dx x log a (4) Differentiation of inverse trigonometrical functions : The following formulae can be applied directly while differentiating inverse trigonometrical functions d 1 , for  1  x  1 sin 1 x  dx 1  x2 (ii) d 1 , for  1  x  1 cos 1 x  dx 1  x2 d 1 , for | x |  1 sec 1 x  dx | x | x2 1 (iii) (iv) d 1 , for x  R tan 1 x  dx 1  x2 (v) d sec h x   sec h x tan h x dx (vii) d sin h1 x  1 / (1  x 2 ) dx (iv) d cot h x   cosec h 2 x dx (vi) D YG (v) d cos h x  sin h x dx ID d tan h x  sec h 2 x dx (iii) (ii) U d sin h x  cos h x dx (ix) d tan h1 x  1 /(x 2  1) dx (xi) d sec h 1 x  1 / x (1  x 2 ) dx d cosec h x  cosec h x cot h x dx (viii) (x) d cos h1 x  1 / (x 2  1) dx d cot h1 x  1 /(1  x 2 ) dx (xii) d cosec h 1 x  1 / x (1  x 2 ) dx U (6) Differentiation by inverse trigonometrical substitution: substitutions following formulae and substitution should be remembered (i) sin 1 x  cos 1 x   / 2 ST (iii) sec 1 x  co sec 1 x   / 2 for d 1 , for x  R cot 1 x  dx 1  x2 (vi) (5) Differentiation of hyperbolic functions : (i) d 1 , cosec 1 x  dx | x | x2 1 E3 | x | 1 60 (i) For trigonometrical (ii) tan 1 x  cot 1 x   / 2 (iv) sin 1 x  sin 1 y  sin 1  x 1  y 2  y 1  x 2     x y  (v) cos 1 x  cos 1 y  cos 1  xy  (1  x 2 )(1  y 2 )  (vi) tan 1 x  tan 1 y  tan 1      1  xy  (vii) 2 sin 1 x  sin 1 (2 x 1  x 2 ) (viii) 2 cos 1 x  cos 1 (2 x 2  1) 2  2x   1  1  x 1  2 x  (ix) 2 tan 1 x  tan 1   cos  sin    1  x 2 1  x 2  1  x 2       (x) 3 sin 1 x  sin 1 (3 x  4 x 3 ) (xi) 3 cos 1 x  cos 1 (4 x 3  3 x )  3x  x3   (xii) 3 tan 1 x  tan 1  2  1  3 x    x  y  z  xyz   (xiii) tan 1 x  tan 1 y  tan 1 z  tan 1   1  xy  yz  zx  132 Differentiation (xv) cos 1 ( x )    cos 1 x (xiv) sin 1 ( x )   sin 1 x (xvi) tan 1 ( x )   tan 1 x or   tan 1 x (xvii)  1  x   tan 1 x  tan 1   4 1  x  (7) Some suitable substitutions Function (i) a2  x 2 x  a sin  or a cos  (ii) x 2  a2 (iii) x 2  a2 x  a sec  or a cosec (iv) ax ax (v) a2  x 2 a2  x 2 x 2  a 2 cos 2 (vi) ax  x 2 x ax x  a tan 2  (x  a)( x  b) (viii) x  a sec 2   b tan 2  3.3 Theorems for Differentiation. Function (x) Substitution x  a tan  or a cot  x  a cos 2 x  a sin 2  E3 (ix) S. N. x  a sin 2  x ax (x  a)(b  x ) ID (vii) Substitution 60 S. N. x  a cos 2   b sin 2  U Let f (x ), g(x ) and u(x ) be differentiable functions D YG (1) If at all points of a certain interval. f (x )  0, then the function f (x ) has a constant value within this interval. U (2) Chain rule (i) Case I : If y is a function of u and u is a function of x, then derivative of y with respect to dy dy du dy du x is or y  f (u)    f ' (u) dx du dx dx dx (ii) Case II : If y and x both are expressed in terms of t, y and x both are differentiable with dy dy / dt  respect to t then. dx dx / dt (3) Sum and difference rule : Using linear property d d d ( f (x )  g(x ))  ( f (x ))  (g(x )) dx dx dx d d d (ii) ( f (x )g (x ))  f (x ) g(x )  g(x ) f (x ) dx dx dx d dw du dv (u.v.w.)  u.v.  v.w.  u.w. dx dx dx dx d d (5) Scalar multiple rule : (k f (x ))  k f (x ) dx dx d d g(x ) ( f (x ))  f (x ) (g(x )) d  f (x )  dx dx   (6) Quotient rule : , provided g( x )  0 dx  g(x )  (g(x ))2 ST (4) Product rule : (i) Example: 5 The derivative of f ( x ) | x | 3 at x  0 is (a) 0 (b) 1 [Rajasthan PET 2001; Haryana CEE 2002] (c) –1 (d) Not defined Differentiation 133 Solution: (a)  x 3 f (x )   3  x , x 0 , x 0 and  3 x 2 f  (x )    3 x 2 , x 0 , x 0 f (0  )  f (0  )  0 The first derivative of the function ( sin 2 x cos 2 x cos 3 x  log 2 2 x  3 ) with respect to x at x   is (a) 2 Solution: (b) (c)  2  2 log e 2 (b) –1 f (x )  sin 2 x. cos 2 x. cos 3 x  log 2 2 x  3 , f (x )  1 3 2 Around x  (b) 0 2 , 3 2 2 dy 2  sin  cos , = dx 3 3 3 (a) e (b) 1/e E3 (c) 1 (d) None of these U If f (x ) | log x |, then for x  1, f (x ) equals (a) Solution: (d) [IIT 1985; Rajasthan PET 2000; MP PET 2000; Karnataka 1 1 1 0  log(log x ) 1 log(log x )  f (x )  x x  f (e )  e  f (x )  log x (log x )  2 1 e log x (log x ) 1 x ST Example: 9 (d) None of these 3 1 1   ( 3  1). 2 2 2 If f (x )  log x (log x ), then f (x ) at x  e is CET 2002] 1 ( 3  1) 2 dy  sin x  cos x dx D YG At x  Solution: (b) (c) | cos x |   cos x and | sin x |  sin x  y   cos x  sin x  Example: 8 dy 2 at x  is dx 3 ID If y | cos x |  | sin x | then (a) Solution: (c) 1 1 1 [7 cos 7 x  cos x ]  1 , f (x )  7 cos 7 x  cos x  1 , f ( )  2  1  1. 4 4 4 U Example: 7 2  log e 2 1 1 sin 4 x cos 3 x  (x  3) log 2 2 , f (x )  [sin 7 x  sin x ]  x  3 2 4 Differentiate w.r.t. x, f (x )  (d) 60 Example: 6 (b) 1 | x|  log x , if 0  x  1  f (x )  f (x ) | log x |   if x  1  log x , (c) 1 x (d) None of these  1  x , if 0  x  1.  1  , if x  1  x Clearly f (1 )  1 and f (1 )  1 ,  f (x ) does not exist at x  1 Example: 10 d dx 3 / 4     x x  2  log    equals to e  x 2        (a) 1 (b) x2  1 x2  4 (c) x2  1 x2  4 (d) e x x2  1 x2  4 134 Differentiation      Let y  log e x  x  2   yx  Example: 11 3 / 4   x 2  x    log e  log  x 2  x 2 3/4 3 1 1  3 dy 3 1  2 1    [log( x  2)  log( x  2)]  4  x  2 x  2  dx 4 (x  4 ) dy x2 1  2. dx x  4  y  x2  If x  exp tan 1  2   x  60 Solution: (c)  dy  then equals  dx   [MP PET 2002] (b) x [1  tan (log x )]  sec 2 (log x ) E3 (a) 2 x [1  tan (log x )]  x sec 2 (log x ) (d) 2 x [1  tan (log x )]  sec 2 (log x ) (c) 2 x [1  tan (log x )]  x 2 sec 2 (log x )  Example: 12 x 2      tan(log x )  y  x 2 tan(log x )  x 2  dy sec 2 (log x )  2 x. tan(log x )  x 2.  2x dx x dy dy  2 x tan(log x )  x sec 2 (log x )  2 x   2 x [1  tan(log x )]  x sec 2 (log x ). dx dx U  y  x2  y  x2   log x  tan 1  2   x   ID  y  x2  x  exp tan 1  2   x   x 1     sin 1  x  1  , then dy  If y  sec 1   x 1  x 1 dx     2002] D YG Solution: (a) (a) 0 1 (b) x 1 [UPSEAT (c) 1   x 1         sin 1  x  1  = cos 1  x  1   sin 1  x  1     dy  0 y  sec 1   x 1   x 1   x 1  2  x 1  dx         Example: 13 d  cos x  sin x  tan 1   dx  cos x  sin x  1 2(1  x 2 ) (b) Example: 14 1 1  x2 (c) 1 (d) – 1   d  cos x  sin x  d  tan 1  tan 1  tan   x   1.  dx 4  cos x  sin x  dx    ST Solution: (d)   1 1  sin x  cos x   2  [AISSE 1985, 87; DSSE 1982, 84; MNR 1985; Karnataka CET 2002; Rajasthan PET U (a)  d  2  1x sin cot 1  dx    1 x     equals   [MP PET 2002; EAMCET 1996] (a) 1 Solution: (b) AMU (d) None of these Solution: (a) 2002, 03] 1999; (b) 1 2 (c)  1 2 (d) 1  1  x  Let y  sin 2 cot 1    1  x  Put x  cos     cos 1 x   1  cos   y  sin 2 cot 1    1  cos     1 1  2 2 1  2  = (1  cos  ) = (1  x )   sin cot  tan   y  sin    = cos 2 2 2 2 2 2       Differentiation 135 dy 1  dx 2  dy  5 cos x  12 sin x    If y  cos 1  is equal to  , x   0,  , then 13 dx    2 Solution: (a) (b) – 1  y  cos 1 {cos( x   )}  x   [Rajasthan PET 1997] (b) sin x d cosh 1 x  dx Example: 18 tan 2 2 x  tan 2 x 1  tan 2 x tan x 2 2 = 1 sec x  1 sec x tan x = 2 sec x tan x  sec x. tan x (c) sec 2 x (b) tan 3 x tan x [AMU 2000] (d) sec x tan x (tan 2 x  tan x ) (tan 2 x  tan x ) = tan( 2 x  x ) tan( 2 x  x ) = tan x tan 3 x. (1  tan 2 x tan x ) (1  tan 2 x tan x ) d d [y. cot 3 x ]  [tan x ]  sec 2 x. dx dx  x x  x x If f (x )  cot 1  2  (a) – 1   , then f ' (1) is equal to   (b) 1  x x  x x f (x )  cot 1  2  [Rajasthan PET 2000] (c) log 2 (d)  log 2     U Solution: (a) d cosh 1 (sec x )  dx D YG Let y   x 1 ,  tan 2 2 x  tan 2 x    cot 3 x   2 2   1  tan 2 x tan x   d dx (a) tan 2 x tan x Solution: (c) 1 2 (d) cosec x ID We know that (c) tan x U Example: 17 (  x   is in the first or the second quadrant) d cosh 1 (sec x ) = dx (a) sec x Solution: (a) (d) None of these dy  1. dx  Example: 16 (c) 0 12 5 Let cos  . Then sin  . So, y  cos 1 {cos . cos x  sin . sin x } 13 13 60 (a) 1 E3 Example: 15  tan 2   1   = cot 1 (  cot 2θ ) =   cot 1 (cot 2 ) Put x x  tan  ,  y  f (x )  cot 1   2 tan    ST  y =   2 =   2 tan 1 (x x )  Example: 19 n If y  (1  x )(1  x 2 )(1  x 4 ).......( 1  x 2 ) then (a) 1 Example: 20 dy at x  0 is dx (b) – 1 (c) 0 n 1 n Solution: (a) dy 2 . x x (1  log x )  f (1)  1. dx 1  x 2 x y (1  x ) (1  x ) (1  x 2 ).....( 1  x 2 ) 1  x 2  1 x 1 x  dy  2 n 1. x 2  dx n 1 1 (1  x )  1  x 2 (1  x )2 n 1 ,  At x  0 , dy  2 n 1 0. 1  1  0  1. dx 12   If f (x )  cos x. cos 2 x. cos 4 x. cos 8 x. cos 16 x then f   is 4 (d) None of these 136 Differentiation f (x )  1 (c) 1 2 (d) None of these 2 sin x. cos x. cos 2 x. cos 4 x. cos 8 x. cos 16 x sin 32 x  5 2 sin x 2 sin x  f (x )  1 32 cos 32 x. sin x  cos x. sin 32 x. 32 sin 2 x    f    4 32. 1  2 1.0 2  1   32.   2 60 Solution: (a) (b) 2 2  2. 3.4 Relation between dy/dx and dx/dy. E3 (a) Let x and y be two variables connected by a relation of the form f ( x , y )  0. Let x be a small  y x  y x  1.  1  lim . x 0 x y  x y   lim x 0 dy dy dx y 1 x .. lim  1. So,  1 [ x 0  y 0 ] .  y 0 dx dx / dy dx dy x y U  D YG Now, dy y dx x  lim and.  lim  y 0 dy y dx x 0 x ID change in x and let y be the corresponding change in y. Then 3.5 Methods of Differentiation. (1) Differentiation of implicit functions : If y is expressed entirely in terms of x, then we say that y is an explicit function of x. For example y = sin x, y = ex, y = x2 + x + 1 etc. If y is related to x but can not be conveniently expressed in the form of y  f (x ) but can be expressed in the form f ( x , y )  0 , then we say that y is an implicit function of x. rule 1 : (a) Differentiate each term of f ( x , y )  0 with respect to x. U (i) Working (b) Collect the terms containing dy / dx on one side and the terms not involving dy/dx on the ST other side. (c) Express dy/dx as a function of x or y or both. Note (ii) :  In case of implicit differentiation, dy/dx may contain both x and y. Working rule  f    dy x    2 : If f(x, y)= constant, then dx  f     y  f f and are partial differential coefficients of f (x , y ) with respect to x and y x x respectively. where Differentiation 137 Note :  Partial differential coefficient of f (x , y ) with respect to x means the ordinary differential coefficient of f (x , y ) with respect to x keeping y constant. If xe xy  y  sin 2 x , then at x  0 , (a) – 1 Solution: (c) dy = dx [IIT 1996] (b) – 2 We are given that xe xy (c) 1  y  sin x 2 When x  0 , we get y  0 (d) 2 60 Example: 21 Example: 22 If sin( x  y )  log( x  y ) , then dy  1. dx dy = dx Roorkee 2000] (a) 2 Solution: (d) (b) – 2 sin( x  y )  log( x  y ) [Karnataka CET 1993; Rajasthan PET 1989, 1992; ID Putting, x  0 , y  0 , we get E3  dy  dy  y   2 sin x cos x Differentiating both sides w.r.t. x, we get, e xy  xe xy  x  dx  dx (c) 1  D YG  1   dy  0 cos( x  y)   1 x  y   dx   U dy  1   Differentiating with respect to x, cos( x  y ) 1  dx  x  y  cos( x  y)  (d) – 1 dy   1  dx    dy dy 1  1. 0, for any x and y. So, 1  dx dx x y 1 cos( x  y )  dy f / x x y Trick: It is an implicit function, so    1. 1 dx f / y cos( x  y )  x y Example: 23 If ln( x  y )  2 xy , then y (0 ) = U (a) 1 ln(x  y )  2 xy  (c) 2 (d) 0 (1  dy / dx ) 12 dy 1  2 xy  2 y  dy   2 x  y   1 , at x  0 , y  1.  y (0 )   (x  y ) 1 dx 2 x 2  2 xy  1  dx  2 ST Solution: (a) (b) – 1 [IIT Screening 2004] (2) Logarithmic differentiation : If differentiation of an expression or an equation is done after taking log on both sides, then it is called logarithmic differentiation. This method is useful for the function having following forms. (i) y   f (x ) g( x ) (ii) y  f1 (x ). f2 (x )......... where gi (x )  0 (where i = 1, 2, 3,.....), fi(x) and gi(x) both are g1 (x ).g2 (x )........ differentiable (i) Case I : y  [ f (x ]g( x ) where f (x ) and g(x ) are functions of x. To find the derivative of this type of functions we proceed as follows: 138 Differentiation Let y = [ f (x )]g( x ). Taking logarithm of both the sides, we have log y  g(x ). log f (x) Differentiating with respect to x, we get  g(x ) df (x )  g(x ) df (x ) dy dg(x )  dg(x )   y.  log[ f (x )].  [ f (x )g( x )   log[ f (x )  dx dx  dx   f (x ) dx  f (x ) dx (ii) Case II : y  60  1 dy 1 df (x ) dg(x )  g(x ).  log { f (x )}. y dx f (x ) dx dx f1 (x ). f2 (x ) g1 (x ).g2 (x ) Differentiating with respect to x, we get E3 Taking logarithm of both the sides, we have log y  log[ f1 (x )]  log[ f2 (x )]  log[ g1 (x )]  log[ g 2 (x )] 1 dy f1(x ) f2 (x ) g1 (x ) g 2 (x )     y dx f1 (x ) f2 (x ) g 2 (x ) g 2 (x ) Working rule : (a) To take logarithm of the function If x m y n  2(x  y)m n , the value of (a) x  y dy is dx x y (c) y x [MP PET 2003] (d) x  y x m y n  2(x  y)m n  m log x  n log y  log 2  (m  n) log( x  y ) D YG Solution: (c) (b) (b) To differentiate the function U Example: 24 ID  f (x ) f2(x ) g1 (x ) g2 (x )  f1 (x ). f2 (x )  f1(x ) f2 (x ) g 1 (x ) g 2 (x )  dy  y 1        =  dx g 1 (x ).g 2 (x )  f1 (x ) f2 (x ) g 1 (x ) g 2 (x )   f1 (x ) f2 (x ) g1 (x ) g2 (x )  Differentiating w.r.t. x both sides m n dy m  n  dy  dy y   1 .  x y dx x  y  dx  dx x Example: 25 dy is equal to dx Solution: (a) [IIT 1994; Rajasthan PET (a) (sin x )tan x. (1  sec 2 x. log sin x ) (b) tan x. (sin x )tan x 1. cos x (c) (sin x )tan x. , sec 2 x log sin x (d) tan x. (sin x )tan x 1 U 1996] If y  (sin x )tan x , then Given y  (sin x )tan x ST log y  tan x. log sin x Differentiating w.r.t. x, 1 dy.  tan x. cot x  log sin x. sec 2 x y dx dy  (sin x )tan x [1  log sin x. sec 2 x ]. dx (3) Differentiation of parametric functions : Sometimes x and y are given as functions of a single variable, e.g., x =  (t) , y =  (t) are two functions and t is a variable. In such a case x and y are called parametric functions or parametric equations and t is called the parameter. To dy find in case of parametric functions, we first obtain the relationship between x and y by dx eliminating the parameter t and then we differentiate it with respect to x. But every time it is Differentiation 139 dy can also be obtained by the following dx not convenient to eliminate the parameter. Therefore dy dy / dt  dx dx / dt To prove it, let x and y be the changes in x and y respectively corresponding to a small formula If x  a(cos    sin  ) , y  a(sin    cos  ), 1t 2 and sin y  t 1t (b) Obviously x  cos 1 1 1t 2 2 , then 1t 1  t2 and y  sin 1 If x  (a) (c) x 1  t2 1t 2 and y  2t 1t y x 2 , then (b) Put [MP PET 1994] 1 1  t2 (d) 1 t 1  t2 dy  1. dx dy  dx [Karnataka CET 2000] y x (c) x y (d) x y 1  t2 2t and y  1  t2 1  t2 U Solution: (c) dy = dx D YG  x  tan 1 t and y  tan 1 t  y  x  Example: 28 (d) cosec (c) sec  ID 1 If cos x  (a) – 1 Solution: (d) [DCE 1999] dy dy / d  a[cos    ( sin  )  cos  ]  sin   =   tan . dx dx / d  a[ sin    cos   sin  ]  cos  U Example: 27 dy  dx (b) tan  (a) cos  Solution: (b) E3 y dy lim dy y t 0 t ' (t)   lim   dt  x dx dx x 0 x  ' (t) lim t 0 t dt y y / t Since  , x x / t Example: 26 60 change t in t. t  tan  in both the equations, we get ST Differentiating both the equations, we get Therefore x 1  tan 2  1  tan  2  cos 2 and y 2 tan  1  tan 2   sin 2. dx dy  2 sin 2 and  2 cos 2. d d cos 2 dy x  . sin 2 dx y (4) Differentiation of infinite series : If y is given in the form of infinite series of x and we have to find out (i) If y  2y dy then we remove one or more terms, it does not affect the series dx f (x )  f (x )  f (x ) ....... , then y  dy f (x ) dy dy  ,   f (x )  dx 2 y  1 dx dx f ( x )  y  y 2  f (x )  y 140 Differentiation If y  f (x ) f ( x ) (ii) f ( x ) f ( x )..... then y  f (x )y  log y  y log f (x ) dy y 2 f (x ) 1 dy y. f (x ) dy   log f (x ).  ,  y dx f (x ) dx dx f (x )[1  y log f (x )] Example: 29 1 f (x ) .... If y  x  x  x ........ to  then dy  dx [Rajasthan PET 2002] 2 2y  1 (a) x 2y  1 Solution: (d) y x  x  x ........ to   y  x  y  y 2  x  y  2 y Example: 30 If y  x x , then x (1  y log e x ) (a) x 2 x (1  y log e x ) y  xx dy  y2 dx Example: 31 (b) y 2 x......  y  x y  log e y  y log e x  1 If y  x 2  x  x2  2 xy 2y  x 2 y  x2  Example: 32 If x  e y e (a) Solution: (c) dy dy dy dy 1 (2 y  1)  1  1   dx dx dx dx 2y  1 [DCE 2000] (d) None of these 1 dy 1 dy y    log e    log e x y dx x dx y  dy y  x    dx x dy  dx (b) xy y  x2 (c) xy y  x2 (d) 2x x2 2 y dy 2 xy dy dy 1  y. 2 x  x 2   y 2  x 2y  1  2y  dx dx dx 2 y  x 2 y ST Solution: (a) 1 2y  1 1 x 2 ......  U (a) , then 1 2 (d) (c) xy 2 D YG Solution: (b) dy is dx 1 2y  1 U x..... (c) ID (b) 60 f (x )  dy y f (x )  dx 2 y  f (x ) then E3 1 If y  f (x )  (iii) 1x x y ........to  , then dy is dx (b) 1 x (c) 1x x (d) x 1x x  e yx Taking log both sides, log x  (y  x ) log e  y  x  y  x  log x  dy 1 dy 1 1 x 1   1   dx x dx x x (5) Differentiation of composite function : Suppose function is given in form of fog(x ) or f [g(x )] Differentiation 141 Working rule : Differentiate applying chain rule If f (x ) | x  2 | and g(x )  f ( f (x )) , then for x>20, g (x ) equals (a) –1 Solution: (b) (b) 1 (c) 0 For x > 20, we have f (x ) | x  2 |  x  2 and, g(x )  f ( f (x ))  f (x  2)  x  2  2  x  4  g ( x )  1 (a) 1  x n Solution: (c) 1 1  xn , then g (x ) equals (c) 1  [g(x )]n (b) 1  [ f (x )]n Since g is inverse of f. Therefore, fog (x )  x for all x  d { fog (x )}  1 for all x dx  f (g(x )).g(x )  1  f {g(x )}  E3 If g is inverse of f and f (x )  (d) None of these 1    f (x )  1  x n    1 1 1   n g(x ) g(x ) 1  [g(x )] ID Example: 34 (d) None of these 60 Example: 33 d f [g(x )]  f ' [g(x )].g' (x ) dx  g(x )  1  [g(x )]n U 3.6 Differentiation of a Function with Respect to Another Function. find D YG In this section we will discuss derivative of a function with respect to another function. Let u  f (x ) and v  g(x ) be two functions of x. Then, to find the derivative of f (x ) w.r.t. g(x ) i.e., to du du / dx du we use the following formula  dv dv / dx dv Thus, to find the derivative of f(x) w.r.t. g(x) we first differentiate both w.r.t. x and then divide the derivative of f(x) w.r.t. x by the derivative of g(x) w.r.t. x. The differential coefficient of tan 1 U Example: 35 2x 2x w.r.t. sin 1 is 2 1  x2 1 x [Roorkee 1966; BIT Mesra 1996; Karnataka CET 1994; MP PET 1999; UPSEAT 1999, 2001] (a) 1 Let y1  tan 1 ST Solution: (a) (b) –1 (c) 0 (d) None of these 2x 2x and y 2  sin 1 1  x2 1  x2 Putting x = tan  y1  tan 1 tan 2  2  2 tan 1 x and y 2  sin 1 sin 2  2 tan 1 x Again and dy 2 d 2  [2 tan 1 x ]  dx dx 1  x2 Hence Example: 36 dy 1 d 2  [2 tan 1 x ]  dx dx 1  x2........(i)........(ii) dy 1 1 dy 2    1  x  The first derivative of the function cos 1  sin  x x  with respect to x at x =1 is  2     142 Differentiation 3 4 (b) 0   1 x f (x )  cos 1 cos    2 2   Solution: (a)  f (x )   (c) 1 2 (d)  1 2    x x    1  x  x x  2 2  1 1 1 3.  x x (1  log x )  f (1)    1  4 4 2 2 1 x 60 (a) 3.7 Successive Differentiation or Higher Order Derivatives. Definition and notation : If y is a function of x and is differentiable with respect dy to x, then its derivative can be found which is known as derivative of first order. If the first dx dy derivative is also a differentiable. function, then it can be further differentiated with dx respect to x and this derivative is denoted by d 2 y / dx 2 which is called the second derivative of y E3 (1) d 2y is also differentiable then its derivative is called third derivative dx 2 ID with respect to x further if d 3y d ny th. Similarly n derivative of y is denoted by. All these dx 3 dx n derivatives are called as successive derivative and this process is known as successive differentiation. We also use the following symbols for the successive derivatives of y  f (x ) : Dy, D YG y 1 , y 2 , y 3,........., y n ,...... U of y which is denoted by D 2 y, D 3 y........., D n y,...... (where D  y I , y II , y III........., y n ,...... dy d 2 y d 3 y d ny , , ,....... ,........... dx dx 2 dx 3 dx n d ) dx f (x ), f (x ), f (x ),........., f n (x ),...... If y  f (x ) , then the value of the nth order derivative at x  a is usually denoted by U  d ny    or (y n ) x a or (y n ) x a or f n (a)  dx n    x a (2) nth Derivatives of some standard functions : dn  n  sin( ax  b)  a n sin   ax  b  n 2 dx   ST (i) (a) (ii) (b) dn  n  cos(ax  b)  a n cos   ax  b  n 2 dx   m! dn (ax  b)m  a n (ax  b)m n , where m  n n (m  n)! dx Particular cases : (i) (a) When m = n D n {(ax  b)n }  a n.n ! (ii) When a  1, b  0 , then y  x n  D n (x m )  m (m  1).......(m  n  1)x m n  (b) When m  n, D n {(ax  b)m }  0 (iii) When a = 1, b = 0 and m = n, (iv) When m  1, y  1 (ax  b) m! x m n (m  n)! Differentiation 143 D n (y)  a n (1)(2)(3)........( n)(ax  b)1n then y  x n  an (1)n (1.2.3......n)(ax  b)1 n   D n (x n )  n ! dx n log( ax  b)  (1)n 1 (n  1)!a n (4) (ax  b)n d n (a x )  a x (log a)n n dx d n ax (e )  a n e ax n dx (6) (i) 60 (5) dn d n ax e sin( bx  c)  r n e ax sin(bx  c  n ) n dx b where r  a 2  b 2 ;   tan 1 , a y  e ax sin( bx  c) (ii) (b)  n 2 y (a) n 2 y Solution: (a) y  (x  1  x 2 )n  [AIEEE 2002] (d) 2 x 2 y n 2 n    dy  n(x  1  x )  ( 1  x 2 ) dy  n x  1  x 2     dx dx 1  x2  n 1   x   n 2  x  1  x 2  1      1  x2       d 2y d 2y dy dy x.  n 2 (x  1  x 2 )n  (1  x 2 ) 2  x.  n2y. 2 dx dx dx dx If f (x )  x n , then the value of f (1)  f ' (1) f ' ' (1) f ' ' ' (1) (1)n f n (1) is   ......  1! 2! 3! n! (b) 2 n 1 (a) 2 n Solution: (c)  dy x  n(x  1  x 2 )n 1  1   dx 1  x2  d 2y dy  x. 1  x2  2 dx  1  x 2 dx  (1  x 2 ). Example: 38 (c) y D YG  e ax cos(bx  c)  r n e ax cos(bx  c  n  ) ID n d 2y dy If y   x  1  x 2  , then (1  x 2 ) 2  x is   dx dx dx n U Example: 37 dn E3 (3) an (1)n n ! (ax  b)n 1 f (x )  x  f (1)  1 , f (x )  nx n f (x )  n(n  1)x n2 n 1 (c) 0 [AIEEE 2003] (d) 1  f (1)  n  f (1)  n(n  1) ….. f (1) f (1) (1)n f n (1) ......  1! 2! n! n n(n  1) n(n  1)(n  2) n !  n C 0  n C 1  n C 2  n C 3 ......  (1)n n C n  0. 1   .....  (1)n 1! 2! 3! n! ST U f n (x )  n!  f n (1)  n! ,  f (1)  Example: 39   e   log  2   n  x    tan 1  3  2 log x  , then d y is (n  1) f ( x )  tan If    2  dx n  1  6 log x   log( ex )    1  (a) tan 1 {(log x )n } Solution: (b) (b) 0  log e  log x 2 We have y  tan 1  2  log e  log x (c) 1/2          tan 1  3  2 log x   tan 1  1  2 log x   tan 1  3  2 log x   1  6 log x   1  2 log x   1  6 log x           tan 1 1  tan 1 (2 log x )  tan 1 3  tan 1 (2 log x )  y  tan 1 1  tan 1 3  Example: 40 (d) None of these dy d ny 0 n 0. dx dx If f (x )  (cos x  i sin x ) (cos 3 x  i sin 3 x )..... (cos( 2n  1)x  i sin(2n  1)x ), then f (x ) is equal to 144 Differentiation (b)  n 4 f (x ) (a) n 2 f (x ) Solution: (b) (c)  n 2 f (x ) (d) n 4 f (x ) We have, f (x )  cos( x  3 x ....  (2n  1)x )  i sin( x  3 x  5 x ....  (2n  1)x )  cos n 2 x  i sin n 2 x  f (x )  n 2 (sin n 2 x )  n 2 (i cos n 2 x )  f (x )  n 4 cos n 2 x  n 4 i sin n 2 x 3.8 nth Derivative using Partial fractions. 60  f (x )  n 4 (cos n 2 x  i sin n 2 x )  f (x )  n 4 f (x ) For finding nth derivative of fractional expressions whose numerator and denominator are rational algebraic expression, firstly we resolve them into partial fractions and then we find nth x4 , then for n > 2 the value of yn is equal to x2  3x  2 If y  (a) (1)n n![16 (x  2)n 1  (x  1)n 1 ] (c) n![16 (x  2)n 1  (x  1)n 1 ] x4 x  3x  2 2  x 2  3x  7  (d) None of these 15 x  14 1 16 = x 2  3x  7   (x  1)(x  2) (x  1) (x  2) U y Solution: (a) (b) (1)n n![16 (x  2)n 1  (x  1)n 1 ] ID Example: 41 1. ax  b E3 derivative by using the formula giving the nth derivative of  y n  D n (x 2 )  D n (3 x )  D n (7)  D n [(x  1)1 ]  16 D n [(x  2)1 ] D YG  (1)n n![(x  1)n 1  16 (x  2)n 1 ] = (1)n n![16 (x  2)n 1  (x  1)n 1 ]. 3.9 Differentiation of Determinants. Let (x) = a1 (x ) b1 (x ) a (x ) b1 (x ) a ( x ) b1 ( x ). Then (x )  1  1 a 2 (x ) b 2 (x ) a 2 (x ) b 2 (x ) a2 (x ) b 2 (x ) If we write (x) = | C 1 C 2 C 3 |. Then (x) | C1 C2 C3 |  | C1 C2 C3 |  | C1 C2 C3 | U R1 ST Similarly, if (x)= R 2 , then R3 R1 R1 R1   (x )  R 2  R 2  R 2 R3 R3 R3 Thus, to differentiate a determinant, we differentiate one row (or column) at a time, keeping others unchanged. Example: 42 If fr (x ), gr (x ), hr (x ), r  1,2,3 are polynomials in x such that fr (a)  g r (a)  hr (a), r  1,2,3 and f1 (x ) f2 (x ) f3 ( x ) F(x )  g 1 ( x ) g 2 (x ) g 3 (x ) , then find F ' ( x ) at x  a h1 ( x ) h 2 ( x ) h 3 (x ) (a) 0 Solution: (a) (b) f1 (a)g 2 (a)h3 (a) [IIT 1985] (c) 1 f1 (x ) f2 (x ) f3 (x ) f1' (x ) f2' (x ) f3' (x ) f1 (x ) f2 (x ) f3 (x ) ' ' ' F' ( x )  g 1 (x ) g 2 (x ) g 3 ( x )  g 1 (x ) g 2 (x ) g 3 ( x ) + g 1 ( x ) g 2 ( x ) g 3 ( x ) h1 ( x ) h 2' ( x ) h 3' ( x ) h1 (x ) h 2 (x ) h3 (x ) h1 (x ) h 2 (x ) h3 (x ) (d) None of these Differentiation 145 f1 (a) f2 (a) f3 (a) f1 (a) f2 (a) f3 (a) f1' (a) f2' (a) f3' (a) ' '  F (a)  g 1 (a) g 2 (a) g 3 (a) + g 1 (a) g 2 (a) g 3 (a) + g 1 (a) g 2 (a) g 3 (a) h1 (a) h 2 (a) h 3 (a) h1' (a) h 2' (a) h 3' (a) h1 (a) h 2 (a) h3 (a) =. 0  0  0  0 x3 Let f (x )  6 1 (b) p + p2 (a) p 3 x Given f (x )  6 1 Solution: (d) sin x cos x 1 0 , 2 p p3 (c) p + p3 dx 3 f (x )  dx 3 cos x dx 3 6 1 0 1 p2 p3 dn dx n x n  n !, dn dx n sin x  sin( x  dn n n cos x  cos( x  ) ) and 2 2 dx n 3  3    3! sin x   cos  x   2  2    f (x )  6 1 0 1 p2 p3 D YG We know that d3 sin x ID dx 3 3 d3 (d) Independent of p U d  x3 [IIT 1997] 2nd and 3rd rows are constant, so only 1st row will take part in differentiation d3 60 sin x cos x d3 [ f (x )] at x  0 is 1 0 where p is a constant. Then dx 3 p2 p3 E3 Example: 43 [ fr (a)  gr (a)  hr (a), r  1, 2, 3] Using these results, d3 at x  0 6  6 1 dx 3 1 0 1 0 = 0 i.e., independent of p. 2 p p3 U dx 3 f (x ) d 3 3.10 Differentiation of Integral Function. If g 1 (x ) and g 2 (x ) both functions are defined on [a, b] and differentiable at a point x  (a, b) and ST f (t) is continuous for g (a)  f (t)  g (b) 1 2 Then d dx Example: 44  g2 ( x ) g1 ( x ) f (t)dt  f [g 2 (x )]g 2 (x )  f [g1 (x )]g1 (x ) = f [g 2 (x )] If F(x )   x3 x2 log t dt (x  0 ) , then F ' ( x ) = (a) (9 x 2  4 x ) log x Solution: (a) d d g 2 (x )  f [g 1 (x )] g 1 (x ). dx dx (b) (4 x  9 x 2 ) log x [MP PET 2001] (c) (9 x 2  4 x ) log x (d) None of these Applying formula we get F' (x )  (log x 3 )3 x 2  (log x 2 )2 x = (3 log x )3 x 2  2 x (2 log x ) = 9 x 2 log x  4 x log x = (9 x 2  4 x ) log x. Example: 45 If x   y 0 1 1  4t 2 dt , then d 2y is dx 2 146 Differentiation (a) 2y Solution: (b) x  (b) 4y y 1 0 1  4t 2 dt  (c) 8y (d) 6 y 2 2 dy dx 1 d y d y 4y 4y dy  1  4y2        1  4y2  4y 2 2 2 2 dx dx dy dx dx 1  4y 1  4y2 1  4y 3.11 Leibnitz’s Theorem. 60 G.W. Leibnitz, a German mathematician gave a method for evaluating the nth differential coefficient of the product of two functions. This method is known as Leibnitz’s theorem. Statement of the theorem – If u and v are two functions of x such that their nth derivative exist then Dn (u.v.)  n C 0 (Dn u)v  n C1 Dn1u.Dv  n C 2 Dn2 u.D 2 v ............  n Cr Dnr u.Dr v .........  u.(Dn v). :  The success in finding the nth derivative by this theorem lies in the proper E3 Note Example: 46 If y  x 2 e x , then value of y n is (b) {x 2  2nx  n(n  1)}e x U (a) {x 2  2nx  n(n  1)}e x (c) {x 2  2nx  n(n  1)}e x (d) None of these Applying Leibnitz’s theorem by taking x 2 as second function. We get , D n y  D n (e x. x 2 ) D YG Solution: (b) ID selection of first and second function. Here first function should be selected whose nth derivative can be found by standard formulae. Second function should be such that on successive differentiation, at some stage, it becomes zero so that we need not to write further terms. = n C 0 D n (e x )x 2  n C1 D n 1 (e x ).D(x 2 )  n C 2 D n  2 (e x ).D 2 (x 2 ) ........... = e x. x 2  ne x.2 x  n(n  1) x e.2  0  0 .......... 2! y n  { x 2  2nx  n(n  1)}e x. Example: 47 If y  x 2 log x , then value of y n is (a) Solution: (b) (1)n 1 (n  3)! x n2 (b) (1)n 1 (n  3)!.2 x n2 (c) (1)n 1 (n  2)! x n2 (d) None of these Applying Leibnitz’s theorem by taking x 2 as second function, we get, D n y  D n (log x. x 2 ) U = n C0 Dn (log x ). x 2 n C1 Dn 1 (log x ).D(x 2 ) n C2 Dn  2 (log x )D 2 (x 2 ) ........... (1)n 1 (n  1)! 2 (1)n  2 (n  2)! n(n  1) (1)n  3 (n  3)!. x  n.. 2 x .2  0  0......... 2! xn x n 1 x n2 ST = = (1)n 1 (n  1)! 2n(1)n  2 (n  2)! n(n  1)(1)n  3 (n  3)!   x n2 x n2 x n2 = (1)n 1 (n  3)! (1)n 1 (n  3)!.2  {(n  1)(n  2)  2n(n  2)  n(n  1)} =. n2 x x n2