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WorkDifferentiation , Energy and Power 25 Differentiation Revision...

WorkDifferentiation , Energy and Power 25 Differentiation Revision Points Derivatives of A Function Differentiatiable Function Let y = f(x) be a real valued function of x. A real valued function y = f(x) is said to be Let y be the small increment in y corresponding differentiable or derivative at a point if its to the small increment x in x. dy derivative exist at that point.  y + y = f(x + x)  y = f(x + x) – y dx  y = f(x + x) – f(x) Dividing by x, we get Derivative of Standard Functions y f (x  x)  f (x)  Function Derivative x x k (constant) 0 (zero) Taking limit as x 0 on both sides, we get xn , n  N nxn–1  y  lim  f (x  x)  f (x)     x 0  lim x 0  This result is valid, even if  x   x  nR lim  f (x  x)  f (x)  sin x cos x If the limit x 0   exists, then  x  cos x – sin x the limit is called the derivative of y or f(x) w.r.t. tan x sec2 x cot x – cosec2 x dy x and is denoted by or f '(x) sec x sec x tan x dx cosec x – cosec x cot x  dy  f (x  x)  f (x)  e x ex  f '(x)  lim h 0   dx  x  ax, a > 0, a  1 ax log a If we replaces x by h, we get 1 log x dy lim  f (x  h)  f (x)  x  f '(x)  h0   dx  h  1 If we replace x by a, then we get the derivative loga x x log a of f(x) at x = a  dy   f (a  h)  f (a)  Left Hand Derivative    f '(a)  lim  h 0    dx  x  a  h  Let y = f(x) is a real valued function of x and x = a is any real number, then the limit Leibnit’Z Notation lim  f (a  h)  f (a)  h 0   , if it exists is called the Let y = f(x) be a real valued function of x. Let y  h  be the small increment in y corrresponding to the left hand derivative of f(x) at x = a and is denoted by f '(a  ) or f ' (a)  or L f ' (a).  y  small increment x in x, then lim x 0   , if it  x   f (a  h)  f (a)   f '(a  )  lim h 0   exists is called the derivative of y w.r.t. x and is  h  dy  f (a  h)  f (a)  denoted by  lim h 0   dx  h   Differentiation 26 If left hand derivative f '(a  ) exist, then Derivative of Difference If u and v are differentiable functions of x and  f (x)  f (a)  f '(a  )  lim x a   , where –  < x – a < 0  xa  dy du dv y = u – v, then   dx dx dx Right Hand Derivative If y = f(x) is a real valued function of x and x = a Derivative of Product is any real number, then the limit If u and v are differentiation functions of x and lim  f (a  h)  f (a)  dy dv du h 0   , if it exists is called the y = uv, then u v.  h  dx dx dx right hand derivative of f(x) at x = a and is denoted If y = ku, where k is constant and u is differentiable by f '(a  ) or f '(a)  or Rf '(a). dy du of x, then k.  f (a  h)  f (a)  dx dx  f '(a  )  lim h 0    h  If y = uvw, where u, v and w are differentiable functions of x, then  f (a  h)  f (a)   lim h 0    h  dy dw dv du  uv  uw  vw If right hand derivative f '(a  ) exists, then dx dx dx dx If y = u 1 u 2.... u n, where u 1, u2,...., u n are   f (x)  f (a)  differentiable functions of x, then f '(a )  lim x a    , where 0 < x – a <   x a  dy  1 du1 1 du 2 1 du n  If f '(a  ) and f '(a  ) are equal, then f(x) is said to  (u1 u 2 ....u n )   ....   dx u  1 dx u 2 dx u n dx  be differentiable at x = a. Relationship Between Derivative of Quotient Continuity and Differentiability If u and v are differentiable functions of x and If a real valued function f is finitely differentiable at any point of its domain, then it is necessarily du dv v u u dy dx dx , provided v  0 continuous at that point. y  , then  v dx v2 If a function f(x) is differentiable at point x = a, then it is also continuous at point x = a. Every continuous function need not be Derivative of Composite differentiable. Function Chain Rune If y = f(u) is differentiable function of u = g (x) is Derivative of Sum differentiable function of x, then y = f(g (x)) is a If u and v are differentiable functions of x and dy dy du dy du dv differentiable function of x and  . y = u + v, then   dx du dx dx dx dx If y = u1 + u2 +...... + un, where u1, u2,....., un If y is differentiable function of u 1, u 1 is are functions of x, then differentiable function of ui + 1, for i = 1, 2,...., n – 1 and un is differentiable function of x, then dy du1 du 2 du   ....  n dx dx dx dx dy dy du1 du 2 du    ....  n dx du1 du 2 du 3 dx  Differentiation 27 Derivative of Some Composite Function 2. If y = cos–1 x, – 1  x  1, 0  y  , then , Function Derivative dy 1 (f(x))n n(f(x))n–1 f '(x)  , |x | 1 dx 1  x2 f '(x) f (x) 2 f (x)   3. If y = tan–1 x, x  R,  y  , then , 2 2 1  f '(x) f (x) (f (x)) 2 dy 1  sin f(x) (cos f(x)) f '(x) dx 1  x 2 cos f(x) (– sin f(x)) f '(x) dy 1 tan f(x) (sec2 f(x)) f '(x) 4. If y = cot–1 x, x  R, 0  y  , then  dx 1  x 2 cot f(x) (– cosec2 f(x)) f '(x) sec f(x) (sec f(x) tan f(x)) f '(x)  5. If y = sec–1 x, such that | x |  1, 0  y  , y  , 2 cosec f(x) (– cosec f(x) cot f(x)) f '(x) e f(x) ef(x) f '(x)  1  , for x  1 a f(x) af(x) log a f '(x) dy  x x 2  1 then  dx   1 f '(x) , for x  1 log f(x)  x x 2  1 f (x) f '(x) loga f(x) As graph of sec–1 x is always increasing, then the f (x) log a derivative of sec–1 x is always positive, then Derivative of Inverse Function d (sec 1 x)  1 If y = f(x) is a differentiable function of x such dx x x2  1 that inverse function x = f–1 (y) exists, then x is differentiable function of y and 6. If y = cosec–1 x, such that dx 1 dy    , where 0 | x |  1,  y , y  0 , then dy dy dx 2 2 dx  1 , for x  1 Derivative of Inverse dy  x x2  1  Trigonometric Functions dx  1 , for x  1  x x2  1   1. If y = sin–1 x, – 1  x  1,  y  , then 2 2 As graph of cosec–1 x is always decreasing, then dy 1 the derivative of cosec–1 x, is always negative,  , |x | 1 dx 1  x2 d 1 then (cos ec 1x)  dx x x2  1  Differentiation 28 Derivative of Composite Functions 1 1 19. cot 1    tan 1 x 20. sec 1    cos 1 x Function Derivative x x f '(x) 1 sin–1 f(x) , | f (x) |  1 21. cosec 1    sin 1 x 1  (f (x)) 2 x   f '(x) 22. sin 1 x  cos 1 x  cos–1 f(x) , | f (x) |  1 2 1  (f (x)) 2  23. tan 1 x  cot 1 x  f '(x) 2 tan–1 f(x) 1  (f (x))2  24. sec 1 x  cos ec 1 x  2  f '(x) cot–1 f(x) 1  (f (x))2  xy  25. tan 1 x  tan 1 y  tan 1    1  xy  f '(x) sec–1 f(x) , | f (x) |  1 f (x) (f (x))2  1  xy  26. tan 1 x  tan 1 y  tan 1    1  xy   f '(x) cosec–1 f(x) , | f (x) |  1 f (x) (f (x))2  1 Important Substitutions Expression Substitution Properties of Inverse Trigonometric Function x = a sin  or x = a cos  a2  x2 1. sin (sin x) = x –1 2. cos–1 (cos x) = x a2  x2 x = a tan  or x = a cot  3. tan (tan x) = x –1 4. cot–1 (cot x) = x a2  a2 x = a sec  or x = a cosec  5. sec–1 (sec x) = x ax ax 6. cosec–1 (cosec x) = x or x = a cos 2 or x = a cos  ax ax 7. sin (sin x) = x –1 8. cos (cos–1 x) = x ax ax or x = a tan  9. tan (tan x) = x –1 ax ax 10. cot (cot–1 x) = x 11. sec (sec–1 x) = x a2  x2 a2  x2 or x2 = a2 cos 2 or x2 = a2 cos  12. cosec (cosec–1 x) = x a 2  x2 a2  x2 13. sin–1 (– x) = – sin–1 x ax  x 2 x = a sin2  14. cos (– x) =  – cos x –1 –1 15. tan–1 (– x) = – tan–1 x x x = a tan2  ax 1  1  1 16. sin    cosec x x x x = a sin2  ax 17. cos1    sec 1 x 1 x (x  a)(x  b) x = a sec2 – b tan2  1  1  1 18. tan    cot x (a  x)(b  x) x x = a cos2 + b sin2   Differentiation 29 Useful Results dy  (f '(x)) m   f '(x) g '(x) h '(x)    k   m h k n dx  (g(x)) (h(x))   f (x) g(x) h(x)  Expression Substitution Result 1 – x2 x = sin  cos2  Type II : 1 – x2 x = cos  sin2  (f(x))g(x) 1 + x2 x = tan  sec2  Method : 1 + x2 x = cot  cosec2  Let y = (f(x))g(x) x2 – 1 x = sec  tan2  Taking log on both sides, we get x2 – 1 x = cosec  cot2  log y = log (f(x))g(x)  log y = g(x) log (f(x)) 2x 1  x 2 x = sin  sin 2 Diff. both sides w.r.t. x, we get 2x 1  x 2 x = cos  sin 2 1 dy f '(x)  g (x)  g '(x)log (f (x)) y dx f (x) 2x x = tan  sin 2 1  x2 dy  f '(x)  1 – 2x2 x = sin  cos 2   y  g(x)  g '(x) log (f (x))  dx  f (x)  2x2 – 1 x = cos  cos 2 1  x2 dy  f '(x)    (f (x)) g(x )  g(x)  g '(x) log (f (x))  x = tan  cos 2 dx  f (x)  1  x2 Type III : 2x x = tan  tan 2 (f 1(x))g1(x) + (f2(x))g2(x) 1  x2 Let y = (f1(x))g1(x) + (f2(x))g2(x) 3x – 4x3 x = sin  sin 3 Let u = (f1(x))g1(x) and v = (f2(x))g2(x) 4x3 – 3x x = cos  cos 3 y=u+v 3x  x 3 Diff. both sides w.r.t. x, we get x = tan  tan 3 1  3x 2 dy du dv   dx dx dx Logarithmic Differentiation Type I : Derivative of Implicit Function m (f (x)) When a function is written in the form y = f(x), it , m, n and k are constant is said to be in explicit form. (g(x)) n (h(x)) k If functions inolving relationship between x and (f (x)) m y such as ax2 + 2hxy + by2 = 0, ex + y = log (x + y) Let y  (g(x)) n (h(x)) k which can not be written in explicit form are called implicit functions. Takign log on both sides, we get If the variables x and y are connnected by a relation  (f (x))m  of the form f(x, y) = 0 and it is not possible or log y  log  n k  convenient to express y as a function of x in the  (g(x)) (h(x))  form y  (x) , then y is said to be an implicit  log y = m log (f(x)) – n log (g(x)) – k log (h(x)) function. Diff. both sides w.r.t. x, we get To find the derivative of implicit function, we 1 dy f '(x) g '(x) h '(x) differentiable each term w.r.t. x. In that case a m h k term containing x is differentiated in ordinary y dx f (x) g(x) h(x) manner, while the term containing y is first dy  f '(x) g'(x) h '(x)  dy   ym h k differentiated w.r.t. y and then multiplied by. dx  f (x) g(x) h(x)  dx  Differentiation 30 e.g. d (y 2 )  2y dy If d2y  f ''(x) is a differentiable function of x, dx dx dx 2 Then we collect all terms containing dy on one d  d2 y  d then its derivative  2  or (f ''(x)) is dx dx  dx  dx side, while teh remaining terms on other side to called the third order derivative of y or f(x) w.r.t. dy 3 obtain. x and is denoted by d y or f '''(x). dx dx 3 In general, the nth order derivative of y = f(x) is Derivative of Parametric Functions n If the relation between x and y is given in terms denoted by d y  f n (x). of another variable say t i.e. x = f(t) and y = g(t), dx n then t is called parametric functions. All these derivatives are called higher order derivatives. dy To obtain in case of parametric functions, These higher order derivatives are denoted by dx first obtain relationship between x and y by y1, y2, y3,...., yn, or y ', y '', y ''',..... y ( n ) or eliminating the parameter t and then differentiable Dy, D2y, D3y,.... Dn y. it w.r.t. x. But every time it is not possible to dy Important Tips and Short Cut Methods eliminate the parameter t. Then obtain by dx d 1. (x)  1 dy dx dy dt using . dx dx d 1 2. ( x)  dt dx 2 x If x = f(t) and y = g(t) are two differentiable d  1  1 functions of t, such that y is defined as function 3.    dx  x  2x x dy dy dt dx d of x, then  , 0. 4. (log (sin x))  cot x dx dx dt dx dt d 5. (log (cos x))   cot x dx Higher Order Derivatives d If y = f(x) is a differentiable function of x, then 6. (log tan x)  2 cos ec 2x dx dy  f '(x) is the first order derivative of y or d dx 7. (log cot x)   2cosec 2x f(x) w.r.t. x. dx dy d If  f '(x) is a differentiable function of x, 8. (log sec x)  tan x dx dx d  dy  d d then its derivative   or (f '(x)) is 9. (log cosec x)   cot x dx  dx  dx dx called the second order derivative of y or f(x) d  1  n 10.   2 d y dx  x n  x n 1 w.r.t. x and is denoted by or f ''(x). dx 2  Differentiation 31 11. If at all points of a certain interval, f '(x)  0 , 16. If f(x) and g(x) are differentiable functions of x, then the function f(x) has a constant value within then derivatives of f(x) w.r.t. g (x) is this interval f '(x) , g '(x)  0. g'(x) d x 12. (x )  x x (1  log x) dx dy y 17. If xmyn = (x + y)m + n, m, n  R, then  and dx x 13. If y  f (x)  f (x)  f (x) ........  , d2y 0. dy f '(x) dx 2 then  dx 2y  1 18. If x = f(t), y = g(t) are differentiable function of t, f ( x )f ( x ).....  d 2 y f '(t).g(t)  g '(t).f ''(t) 14. If y  f (x)f (x ) , then then  dx 2 (f '(t))2 dy y 2 f '(x)  dy y dx f (x)(1  y log f (x)) 19. If ax2 + 2hxy + by2 = 0, then  and dx x 1 d2 y 15. If y  f (x)  1 , 0 f (x)  dx 2 f (x) ....  dy yf '(x) then  dx 2y  f (x)  Differentiation 32 Multiple Choice Questions Differentiability 1  x, for x  2 6. If f (x)   , then: 1. Which of the following is not true ? 5  x, for x  0 A) A continuous function is always A) f(x) is continuous and differentiable at x = 2 differentiable B) f(x) is continuous at x = 2 but not B) A differentiable function is always continous differentiable C) A polynomial function is always continous C) f(x) is differentiable at x = 2 D) ex is continuous for all x D) f(x) is not continuous x = 2 2. Which of the following is not true always ?  x  1, for x  2 7. If f (x)   , then at x  2 A) If f(x) is continuous at x = a, then it is 2x  3, for x  0 differentiable at x = a A) f(x) is continuous but not differentiable B) If f(x) is not continuous at x = a, then it is B) f(x) is continuous and differentiable not differentiable at x = a C) f(x) is differentiable C) If f(x) and g(x) are differentiable at x = a, D) f(x) is not continuous then f(x) + g(x) is also differentiable at x = a  x, for 0  x  1 D) If f(x) is continuous at x = a, then lim x a f (x) 8. If f (x)   , then: 2x  1, for 1  0 exists A) f(x) is discontinuous at x = 1  x, for x  0 B) f(x) is differentiable at x = 1 3. If f (x)   , then :   x, for x  0 C) f(x) is continuous but not differentiable at x=1 A) f(x) is continuous at x = 0 but not D) none of these differentiable B) f(x) is continuous and differentiable at x = 0  x  2, for  1 x  3  C) f(x) is differentiable at x = 0 9. If f (x)   5,for x  3 , then at x = 3,  8  x,for x  3 D) f(x) is not continuous at x = 0  f '(x)   x, for x  0 4. If f (x)   , then at x  0 : A) –1 B) 0  0, for x  0 C) 1 D) does not exist A) f(x) is differentiable x B) f(x) is not continuous 10. The set of points where f (x)  is 1 | x | C) f(x) is continuous but not differentiable differentiable is : D) f(x) is not continuous and differentiable A) (  , 0)  (0,  )  x2 B) (  ,  1)  ( 1, )  ,for x  0 5. If f (x)   | x | , then: C) (0, )  0, for x  0  D) (  ,  ) A) f '(x) exists in (– 2, 2) 11. If f(x) = | x – 1 |, x  R , then at x = 1 : B) f '(x) exists in (– 1, 1) A) f(x) is differentiable B) f(x) is not continuous C) f(x) is discontinuous every where C) f(x) is continuous but not differentiable D) f(x) is continuous every where D) f(x) is continuous and differentiable  Differentiation 33 12. If f(x) = | x – 1| + |x + 2|, then at x = 2 :  p 1 A) f(x) is differentiable  x sin   ,for x  0 18. Let f (x)   x , then f(x) is B) f(x) is not differentiable  0, for x  0  C) f '(2  )  2 continuous but not differentiable at x = 0, if A) 1 p   B) 0  p 1 D) f '(2  )  0 C)   p  0 D) p=0  1  x cos   , for x  0  e x  ax, for x  0 13. If f (x)   x , then at x  0 19. If f (x)   is differentiable at  b(x  1) , for x  0 2 0, for x  0  x = 0, then (a, b) is A) f(x) is not continuous A) (– 3, – 1) B) (3, 1) B) f(x) is differentiable C) (– 3, 1) D) (3, – 1) C) f(x) is continuous and differentiable 20. If f(x) = ae |x| + b |x|2, a, b R and f(x) is D) f(x) is continuous but not differentiable differentiable at x = 0, then : A) a = 1, b = 3 B) a = 0, b R    1, for 2  x  0 C) a = 1, b = 2 D) a = 2, b = 3 14. If f (x)   , then at x  0 1  sin x,for 0  x    2 Derivatives of Composite Functions lim  f (x)  f (2)  f '(x)  21. If x 2   exist, then  x2  A) 1 B) –1 A) lim x 2 f (x)  lim x 2 f (x) C)  D) does not exist B) f(x) is differentiable 15. If f (x)  [x],  2  x  2, then at x  1 C) lim x 2 f (x)  f (2) A) f(x) is continuous and differentiable B) f(x) is not continuous and not differentiable D) lim x 2 f (x)  f (2) C) f(x) is continuous 22. If f is derivable at x = a, then D) f(x) is differentiable  xf (a)  af (x)   lim x a  16. If f(x) is derivative at x = 2, where  xa  A) af '(a)  f (a) B) f (a)  af '(a)  x 2 ,for x  2 f (x)   , then ax  b, for x  2 C) af(a) D) af '(a) 23. Derivative of even function is : A) a = 4, b = – 4 B) a = – 4, b = 4 A) non-negative B) even function C) a = 4, b = 4 D) a = – 4, b = – 4 C) odd function D) either A or B 17. The value of m for which the function 24. Derivative of odd function is : A) negative B) even function  mx 2 , for x  1 f (x)   is differentiable at x = 1, C) odd D) odd function  2x,for x  1 25. I f(x) is polynomial of degree two and f(0) = 4, is f '(0)  3 , f ''(0)  4 , then f(– 1) = A) 0 B) 1 A) 2 B) –2 C) 2 D) does not exist C) 3 D) –3  Differentiation 34 dy 5 3 26. If y = (5x3 – 4x2 – 8x)9, then  C) (2x 2  7x  4) 2 (2x  7) dx 3 A) 9 (5x3 – 4x2 – 8x)8 (15x2 – 8x – 8) 3 B) 9 (5x3 – 4x2 – 8x)9 (15x2 – 8x – 8) 5 D) (2x 2  7x  4) 2 (4x  7) 3 C) 9 (5x3 – 4x2 – 8x)8 (5x2 – 8x – 8) D) 9 (5x3 – 4x2 – 8x)9 (5x2 – 8x – 8) dy 31. If y  x  x  x , then  dx 1 dy 27. If y , then  (x 2  3)2 dx  1  1 1  A) 1  1    x x  x    2x 2x x x x  A) B) (x 2  3)3 (x  3)3 2 1  1 1   4x 4x B) 1  1    x x  x   C) D) 2 x x x  (x 2  3)3 (x  3)3 2 1 dy 1  1  1  28. If y  x  , then  C) 1   1   x dx  2 x x  x   2 x x x  x2  1 1  x2 A) B) 1  1  1  2x 2 x 2  1 2x 2 x 2  1 D) 1  1     2 x x  2 x  2 x x x  x2  1 1  x2 C) D) 1 x dy 2x x x2  1 2x x x2  1 32. If y  , then (1  x 2 ) y 1 x dx  1  5 dy A) 1 B) –1 29. If y   x   , then dx  C) 2 D) 0  x 1 1 dy 5(x  1)  1  4 33. If y   , then  A)  x  3x  7 7  3x dx 2x x  x 4   B) 5(1  x)  1  A) 3 1 1   x   2x x  x  2 3 3   (3x  7) 2 (7  3x) 2  4 C) 5(x  1)  1   x  3  1 1  2 x  x B)   2  3 3  4  (3x  7) 2 (7  3x) 2  D) 5(1  x)  1   x  2 x  x   3 1 1  dy C)  30. If y  3 (2x  7x  4) , then 2 5   2 3 3  dx  (3x  7) 2 (7  3x) 2  3 5 A) (2x 2  7x  4) 2 (2x  7)   3 3 1 1  D)   2  3 3   (3x  7) (7  3x) 2 2 5 2  B) (2x 2  7x  4) (4x  7) 3 3  Differentiation 35 x2 x3 dy dy 34. If y  1  x   .... , then  40. If y = sin (cos (tan x)), then  2! 3! dx dx A) – sin (cos (tan x)) sin (tan x) sec2 x A) y B) –y B) sin (cos (tan x)) sin (tan x) sec2 x C) 1 D) 0 C) – cos (cos (tan x)) sin (tan x) sec2 x dy D) cos (cos (tan x)) sin (tan x) sec2 x 35. If y  a  a  x , then  2 2 2 dx dy 41. If y  sin sin x , then  ax dx A) 4 a 2  x2 a2  a2  x2 cos x cos sin x A) ax 2 x sin x B) 2 a 2  x2 a2  a2  x2 cos x cos sin x B) x 2 sin x C) 4 a  x 2 2 a  a x 2 2 2 cos x cos sin x x C) D) 4 x sin x 2 a 2  x2 a2  a2  x2 dy cos x cos sin x  D) 36. If y = sin (x2 + x), then dx 4 sin x A) – (2x + 1) cos (x + x)2 dy B) (2x + 1) cos (x2 + x) 42. If y  cos x , then  dx C) – (2x + 1) sin (x2 + x) D) (2x + 1) sin (x2 + x) sin x  sin x A) B) dy 2 x 2 x 37. If y = sin (x2 + 5), then  dx sin x  sin x A) x cos (x2 + 5) C) D) 2x x 2x x B) (x + 5) cos (x2 + 5) C) 2x cos (x2 + 5) dy 43. If y = cos (2x + 450), then  D) (2x + 5) cos (x + 5) 2 dx A) 2sin (2x + 450) dy 38. If y = sin (ax2 + bx + c), then  B) – 2sin (2x + 450) dx A) (2ax + b) cos (ax2 + bx + c)  C) sin (2x  450 ) B) (2ax + b) sin (ax2 + bx + c) 90 C) (ax + 2b) cos (ax2 + bx + c)  D) sin (2x  450 ) D) (ax + 2b) sin (ax + bx + c) 2 90 dy dy 39. If y = sin (x2 + x + 5), then  44. If y = cos (sin x), then  dx dx A) – (x + 1) cos (x2 + x + 5) A) (sin x) sin (sin x) B) (x + 1) cos (x2 + x + 5) B) – (sin x) sin (sin x) C) – (2x + 1) cos (x + x + 5)2 C) (cos x) sin (sin x) D) (2x + 1) cos (x2 + x + 5) D) – (cos x) sin (sin x)  Differentiation 36 dy dy 45. If y = cos (x2 ex), then  50. If y = sin2 (log (2x + 3)), then  dx dx A) ex (x2 + 2) sin (x2 ex)  sin (2 log (2x  3)) A) B) – ex (x2 + 2) sin (x2 ex) 2x  3 C) xex (x + 2) sin (x2 ex) B) sin ( log (4x  6)) D) – xex (x + 2) sin (x2 ex) 2x  3 1 dy 2sin ( log (4x  6)) 46. If y  , then  C) cot (a tan x  b cot x) dx 2x  3 A) (asec2 x + bcosec2) sec2 (atan x + bcot x) 2sin ( 2 log (2x  3)) B) (asec2 x – bcosec2) sec2 (atan x + bcot x) D) 2x  3 C) (asec2 x – bcosec2) cosec2 (atan x + bcot x) 2 D) – (asec2 x – bcosec2) cosec2 (atan x + bcot x) dy 51. If y  sin 5 (log x), then  dx dy 47. If y  sec (tan x ), then  2 3 dx A) cos (log x) sin 5 (log x) 5x tan x sec2 x sec(tan x ) A) 3 2 x 2 B) cos (log x) sin 5 (log x) 5x sec 2 x sec(tan x ) tan (tan x ) B) 2 2 2 x C) sin (log x) sin 5 (log x) 5x sec 2 x sec (tan x ) tan (tan x ) 2 2 C) D) sin (log x) sin 5 (log x) x 5x tan x sec2 x sec(tan x ) dy D) 52. If y = cos2 (log (2x + 3)), then  dx x A) 2sin (2 log (2x  3)) dy 48. If y  sin x , then  2x  3 dx  2sin (2 log (2x  3)) B) cos x  cos x 2x  3 A) B) 2 sin x 2 sin x C) sin (2 log (2x  3)) cos x  cos x 2x  3 C) D) sin x sin x  sin (2 log (2x  3)) D) 2x  3 dy 49. If y  sin x , then  dx dy 53. If y  tan x , then  dx cos x cos x A) B) sec2 x sec2 x 2 sin x 4 sin x A) B) 4 x tan x 2 x tan x cos x cos x C) D) sec2 x sec2 x 2 x sin x 4 x sin x C) D) 4 x tan x 2 x tan x  Differentiation 37 dy  54. If y  tan (log x ), then  2 2 3 dx 58. If r  2 cos  2    , then  4 A) 2 tan (log x 3 ) sec 2 (log x 3 )  dr x at   ,  4 d B) 3tan (log x 3 ) sec 2 (log x 3 ) 1 1 x  1 2  2 2 A) 2  B) 2    1   1 6 tan (log x 3 ) sec 2 (log x 3 ) C) 1 1 x  1 2  2 2 C) 2  D) 2  2 tan (log x 3 ) sec 2 (log x 3 )   1   1 D) 3x dy 59. If y  cos x  cos x , then  d dx 55. (cosecx  cot x)  dx  sin x sin x A)  A) cosec x (cosec x – cot x) 2 cos x 4 x cos x B) – cosec x (cosec x – cot x) C) cosec x (cosec x + cot x)  sin x sin x B)  D) – cosec x (cosec x + cot x) 2 cos x 2 x cos x dy 56. If y  sin x  cos x , then  sin x sin x dx C)  2 cos x 2 x cos x 2cos x cos x  sin x A) sin x sin x 4 cos x sin x  cos x D)  2 cos x 4 x cos x 2 cos x cos x  sin x B) 60. If y  sin 3 (5x  3)  cos3 (5x  3), then 4 cos x sin x  cos x dy  2cos x cos x  sin x dx C) 2 cos x sin x  cos x 3 A) cos (5x  3) sin sin (5x  3) 2 2 cos x cos x  sin x D) 3 2 cos x sin x  cos x  sin (5x  3) cos(5x  3) 2 1 dy 3 57. If y  , then  B) cos (5x  3) sin sin (5x  3) 3 cosecx  cot x dx 2 3 4cosec x  sin (5x  3) cos(5x  3) A) 2 3 cos ec x  cot x 3 15 C) cos (5x  3) sin sin (5x  3)  4 cosec x 2 B) 3 3 cos ec x  cot x 15  sin (5x  3) cos(5x  3) 2 cosec x C) 15 3 cos ec x  cot x 3 D) cos (5x  3) sin sin (5x  3) 2  cos ec x 15 D) 3 cos ec x  cot x 3  sin (5x  3) cos(5x  3) 2  Differentiation 38 dy cos x  sin x dy 61. If y = tan (2x0) – sin (3x0), then  66. If y  , then  dx cos x  sin x dx 2 2   A) B) A) sec2 (2x 0 )  cos (3x 0 ) 1  sin 2x 1  sin 2x 90 60 1 1 C) D)   1  sin 2x 1  sin 2x B) sec2 (2x 0 )  cos (3x 0 ) 90 90 sin 2 x dy 67. If y  , then    1  cos x 2 dx C) sec2 (2x 0 )  cos (3x 0 ) 60 60  sin 2x sin 2x A) B) (1  cos2 x)2 (1  cos2 x)2   D) sec 2 (2x 0 )  cos (3x 0 ) 180 180  2sin 2x 2sin 2x C) D) (1  cos2 x)2 (1  cos2 x)2 62. If f(x) = cos x cos 2x cos 4x cos 8x cos 16x, then sec x 0  tan x 0 dy  68. If y  , then  f '   sec x  tan x 0 0 dx 4    x0    x0  1 3 A) tan    sec 2    A) B) 180 4 2  4 2  2 2    x0    x0  C) 2 D) 1 B) tan    sec 2    180 4 2  4 2  dy 63. If y = cos (x3) sin2 (x5), then     x0    x0  dx C) tan    sec 2    360 4 2  4 2  A) 5x4 sin (x3) sin (2x5) – 3x2 cos (x3) sin (x5) B) 5x4 sin (x3) sin (2x5) + 3x2 cos (x3) sin (x5)    x0    x0  D) tan    sec 2    C) 5x4 cos (x3) sin (2x5) – 3x2 cos (x3) sin (x5) 90 4 2  4 2  D) 5x4 cos (x3) sin (2x5) + 3x2 cos (x3) sin (x5) 1  sin x dy 69. If y  , then  dy 1  sin x dx 64. If y = sin2 3x tan3 2x, then  dx 1 1 A) B) 1  sin x A) sin 3x tan 2x (2sin 3x sec 2x + 3cos 3x 2 2 1  sin x tan 3x) 2 2 B) 2sin 3x tan2 2x (sin 3x sec2 2x + 3cos 3x C) 1  sin x D) 1  sin x tan 3x) C) 3sin 3x tan2 2x (2sin 3x sec2 2x + cos 3x 1  sin10x dy 70. If y  , then  tan 3x) 1  sin10x dx D) 6sin 3x tan2 2x (sin 3x sec2 2x + cos 3x tan 3x)   A) 5sec 2   5x  m n nq qm 4   sin m x   sin n x   sin q x  65. If y   n   q   m  ,    sin x   sin x   sin x  B)  5sec 2   5x  4  dy   then  C) 5sec 2   5x  dx 4  A) sin x B) cos x   C) 1 D) 0 D) 5sec 2   5x   4   Differentiation 39 dy 1  tan 2x 76. If y  e , then  log x dy 71. If y  , then  dx 1  tan 2x dx 1 A) elog x B)   1  tan 2x x A)  sec 2   2x  4  1  tan 2x C) 0 D) 1 dy 77. If y  e  log (log x )   1  tan 2x , then B) sec   2x  2 dx 4  1  tan 2x 1 1 A) B)   1  tan 2x x log x C)  sec 2   2x  4  1  tan 2x 1 x C) D) x log x log x   1  tan 2x D) sec 2   2x  4  1  tan 2x x 1 dy 78. If y  7 x , then  dx 2x 3 dy 72. If y  e5x , then  dx x 1  x2  1  A) 7 x (log 7)  2  5x 2  x  3 5x 2  x  3  x  A) (5x  1)e B) (5x 2  1)e x 1  x2  1  C) e5x 2 x3 D) e5x 2  x 3 B) 7 x (log 7)  2  (10x  1)2 (10x  1)  x  2 dy 1  x2  1  73. If y  e  cos ec x x , then C) 7 x dx  2   x  2 A) 2ecos ec x cosec x cot x x 1  x2  1  2 D) 7 x  2  B)  2ecos ec x cos ec x cot x  x  C) 2 2ecos ec x cosec 2 x cot x log (1 log x ) dy 79. If y  a , then  dx 2 D)  2ecos ec x cosec 2 x cot x a log (1 log x ) log a a log (1 log x ) log a A) B) dy x  log x x (1  log x) x sin 2x  cos 2 x 74. If y  e , then  dx a log (1 log x ) log a a log (1 log x ) log a C) D) A) (2x cos 2x – sin 2x) exsin2x + cos 2x 1  log x 1  x log x B) (2x cos 2x + sin 2x) exsin2x + cos 2x dy 80. If y  a  x sin x C) (2x cos 2x – sin 2x) exsin2x + cos 2x then dx D) (2cos 2x + xsin 2x) exsin2x + cos 2x  sin x  A) a x sin x (log a)  x cos x   2 2 sin x  cos 2 dy  x  75. If y  e  x x , then dx  sin x  x 2 sin 2 x  cos 2 x B) a x sin x (log a)  x cos x   A) (2xsin x – (x – 1) sin 2x) e 2  x  2 sin 2 x  cos 2 x B) (2xsin2 x + (x – 1) sin 2x) e x  sin x  C) a x sin x (log a)  x cos x   C) (2