Mathematics-14.Unit-10 Differentiation

DIFFERENTIATION SYLLABUS Differentiability of a function at a given point, Differentiation of the sum, difference, product and quotient of two functions, differentiation of trigonometric functions, inverse trigonometric functions, logarithmic functions, exponential functions, composite functions and implicit functions. derivatives of order upto two. DIFFERENTIABILITY OF A FUNCTION AT A GIVEN POINT Let f (x) be a real valued function defined on an open interval (a,b) where c Then f (x) is said to be differentiable or derivable at x = c, exists finitely. This limit is called the derivative or differential coefficient of the function f(x) at x = c, and is denoted by f’(c) or D f (c) or (f (x))x = c Thus, f (x) is differentiable at x = c = = Here = is called the left hand derivative of f (x) at x = c and is denoted by f’ or LF’(c). While , = is called the right hand derivative of f (x) at x=c and is denoted by f’ or Rf’ (c). Thus f (x) is differentiable at x = c. Lf’(c) = Rf’ (c) If Lf’ (c) Rf’(c) we say that f (x) is not differentiable at x = c. Example -1: The set of triplets (a, b, c) of real numbers with a ¹0, for which the function f(x) = , is differentiable, is (A) { ( a, 1- 2a, a) / a ÎR; a ¹ 0 } (B) { ( a, 1- 2a, c) /a, c ÎR; a ¹ 0 } (C) { (a, b, c)/ a, b, c ÎR; a + b+ c = 1 } (D) { ( a, 1- 2a, 0) / a ÎR; a ¹ 0 } Solution: (A) Given f is differentiable for all real x Þ f is continuous for all real x. so, f(x) = f(1) Þ a + b + c =1 . . . . (1) Also f¢(x) = f¢ (1+) = f¢(1–) Þ 1 = 2a + b Þ b = – 2a + 1 . . . . (2) since a, b, c Î R and a ¹ 0, using (1) and (2) Þ c = a DIFFERENTIABILITY IN AN INTERVAL 2.1 IN OPEN INTERVAL A function f(x) defined on an open interval (a, b) is said to be differentiable or derivable in open interval (a, b) if it is differentiable at each point of (a, b) 2.2 IN CLOSE INTERVAL A function f(x) defined on [a, b] is said to be differentiable or derivable at the end points a and b if it is differentiable from the right at a and from the left at b. In other words and both exist. “If f is derivable in the open interval (a, b) and also at the end points a and b, then f is said to be derivable in the closed interval [a, b]”. For checking differentiability on a closed interval [a, b] we say, “A function f is said to be differentiable function if it is differentiable at every point of its domain.” Example -2: Let f(x) = x3 – x2 + x + 1 g(x) = Discuss the continuity and differentiability of g(x) in (0, 2). Solution: f(x) = x3 – x2 + x + 1 Þ f¢(x) = 3x2 – 2x + 1 > 0 " x Þ f(x) is an increasing function on [0, x] Hence, g(x) = Clearly g(x) is continuous at x = 1 and not differentiable at x = 1. DERIVATIVE OF F(X) FROM THE FIRST PRINCIPLE = f (x) The limit which is a function of x is called the derivative of f (x) and is denoted by f (x). The symbol f (c), then denotes the value of the function f (x) for x = c. Example -3: Find the derivative of x3. Solution: Let f (x) = x3 So f (x) = = = = f (x) = 3x2. Example -4: Find the derivative of . Solution: Let f (x) = So, = = = LIST OF DERIVATIVES OF IMPORTANT FUNCTIONS 1. 7. 2. 8. 3. 9. 4. 10. 5. 11. 6. 12. Example -5: Find the derivative of . Solution: Example -6: Find the derivative of . Solution: Example -7: Find the derivative of . Solution: FUNDAMENTAL RULES FOR DIFFERENTIATION (i) Differentiation of a constant function = 0 i.e. (ii) Derivative of the sum: Let u, v be two derivable functions of x so denoting their sum by y, we write, y = u + v so, (iii). Derivative of the difference: Let u, v be two derivable function of x, so denoting their difference by y, so we write y = u –v so, (iv). Generalisation: By a repeated application of the results obtained above, it can be proved that if u1, u2 ……, un be any finite number of derivable functions, then y = u1  u2  u3  …….  un So,  Example -8: Find , if y = . Solution: Let y = x + , so let u = x1 and v = So y = u + v = 1 – = Example -9: Find , if y = . Solution: Let y = x2 – , so let u = x2 and v = So y = u – v = 2x + = Example -10: Find of (a). y = (x + 2)(x + 3) (b). y = (c). y = (d). y = Solution: (a). = (x + 2).1 + (x + 3).1 = x + 2 + x + 3 = 2x + 5 (b). = (c). (d). = DERIVATIVES OF FUNCTIONS IN PARAMETRIC FORMS Sometimes the relation between two variables is neither explicit nor implicit, but some link of a third variable with each of the first two variables, separately, establishes a relation between the first two variables. In such a situation, we say that the relation between them is expressed parametrically. The third variable is called the parameter. More precisely, a relation expressed between two variables x and y in the following form x = f (t), y = g(t) is said to be parametric form with ‘t’ as a parameter. In order to find differentiation of functions in such form, we have by chain rule. Example -11: Find of x = 2at2 , y = 2at3 Solution: , Example -12: If x and y  tan–1 (2t  1), find . Solution: Here x On differentiating both sides, we get,  and y  tan–1 (2t  1) On differentiating both sides, we get,   Hence, DERIVATIVES OF A COMPOSITE FUNCTION Given a composite function y = f(x), i.e. a function represented by y = F(u), u = (x) or y = F [ (x)], then This is called the chain rule. The rule can be extended to any number of composite variables; e.g., if y = f(u(v)), then Example -13: Find dy/dx (i) (ii) (iii) Solution: (i) (ii) Let (iii) = DERIVATIVE OF AN IMPLICIT FUNCTION If x and y are related by the rule F (x, y) = 0 such that y cannot be obtained entirely or exactly in terms of x then y is said to be an implicit function of x. For example: Here we do not get a unique value of y for each x. Eg. x2 – y3 + 3x2y = 0 Here also y cannot be obtained entirely in terms of x. To find in such cases start differentiating the given equation as it is (using rule of composite functions) For example: for x3 + y2 = a2, we have 3x2 + 2y = 0 If y can be expressed entirely in terms of x, then y is said to be an explicit function of x. Note that every explicit function can be written as the implicit function y – f(x) = 0. Example -14: Find dy/dx (i) log(xy) = x2 + y2 (ii) x + y = sin (xy) Solution: (i) log (xy) = x2 + y2  logx + logy = x2 + y2 Differentiating w.r.t x (ii) x + y = sin (xy) Differentiating w.r.t. x, we get 1 + = cos (xy). INVERSE FUNCTIONS AND THEIR DERIVATIVES Theorem: If the inverse functions f and g are defined by y = f(x) and x = g(y) and if f’(x) exists and f’(x)  0 then g’(y) = . This result can also be written as, if exists and , then or =1 or Result: 1. 2. 3. 4. 5. 6. Example -15: Find of (a). y = sin 2x (b). y = cos 2x (c). y = x sin–1 x (d). y = a tan–1 x (e). y = sec2 x (f). y = (log2 x)2 Solution: (a). = 2 cos 2x (b). = – 2 sin 2x (c). = x. + sin–1 x (d). = a (e). = 2 sec x . sec x tan x = 2 sec2 x tan x (f). = 2 log2 x  loge2 = 2log2x . log2e HIGHER ORDER DERIVATIVES Let y = f(x) First derivative Second derivative Third derivative etc. Example -16: Find the second derivative of ax3 + bx2 + cx + d. Solution : Let y = ax3 + bx2 + cx + d. Logarithmic Differentiation If u and v are functions of independent variable x then to differentiate the functions like uv , first we take the log and then differentiate. Example -17: Differentiate w.r.t. sin( m cos– 1x). Solution: Let y = and z = sin(m cos– 1x) So logy = tanx log log x and so . Example -18: If xy . yx  1, find . Solution: Taking log on both sides; Y log x x log y  log 1 Differentiating both sides, we get, or or Differentiating a function w.r.t. to another function Let we have to differentiate f(x) with respect to g(x). If y = f(x) and t = g(x) then we have to find First we find and then Example -19: Differentiate sin2x w. r. t. (logx)2 Solution: Let y = sin2x and t = (logx)2 Example -20: If x2  y2  xy  2, find Solution: x2  y2  xy  2, Differentiating both sides we get, Example -21: If Solution:   Differentiating both sides, we get   Example -22: If 5f(x)  3f  x  2 and y  xf (x), then find at x  1. Solution: Here, Put x  we get, Solving (i) and (ii) we get,  y  x f (x)  or Now at x  1  Example -23: Find a, b, c and d, where f (x)  (ax  b) cos x  (cx  d) sin x and f’ (x)  x cos x is identity in x. Solution: Here, f’ (x)  x cos x  a cos x (ax  b) sin x  c sin x  (cx  d) cos x  x cos x or (a  cx  d) cos x  (ax – b  c) sin x  x cos x  0.sin x  a  b  cx  x and axb  c  0 Which is again identity in ‘x’  a  b  0, c  1, a  o, b  c  o  a 0, b  1, c  1, d  0 SOLVED PROBLEMS SUBJECTIVE 1. The function f(x) = (x2 – 1) |x2 – 3x + 2| + cos (|x|) is not differentiable at : (A) x = –1 (B) x = 0 (C) x = 1 (D) x = 2 Solution : (D) Since cos (–x) = cos x  cos | x | is differentiable for each . Also, Similarly, If , then f (x) is not differentiable at points where g(x) is so. Now, and  g (x) is not differentiable at x = 2 f (x) is not differentiable at x = 2. Hence (D) is correct answer. 2. y = , then is (A) (B) (C) (D) none of these Solution: y = = = (x3  5)1 =  1 (x3  5)2  3x2 = Hence (A) is correct answer. 3. y = esin 2x, then is (A) x cos 2x (B) 2y cos 2x (C) xy cos 2x (D) none of these Solution: y = esin2x  = esin2x cos 2x . 2 = 2 esin2x. cos 2x Hence (B) is correct answer. 4. y = (sinx + cosx)x , then is (A) y(sinx + cosx) (B) y (C) (D) none of these Solution: y = (sin x + cos x)x log y = x log (sin x + cos x) = log (sin x + cos x) + = (sin x + cos x)x Hence (B) is correct answer. 5. y = , then is (A) 5 (B) 2 (C) 0 (D) none of these Solution: y = = sin1 x + cos1 x =  = 0 Hence (C) is correct answer. 6. y = , then is (A) (B) (C) (D) none of these Solution: y = log y = sin1 x log 2 + cos1 (x  2) = answer is none of these. Hence (D) is correct answer. 7. x = a cos3 t, y = a sin3 t , then is (A) – tant (B) – cos t (C) – cosec t (D) – cot t Solution: = 3a sin2 t cos t, =  3a cos2 t sin t =  tan t Hence (A) is correct answer. 8. x = t2 + t + 1, y = t2 – t + 1; then is (A) (B) (C) (D) none of these Solution: x = t2 + t + 1, = 2t + 1, = 2t  1 Hence (B) is correct answer. 9. x = a (1 – sint), y = a(1 – tan t) , then is (A) sec2t (B) sec3t (C) cos2t (D) cos3t Solution: x = a (1  sin t)  = a (0  cos t) =  a cos t = sec3 t Hence (B) is correct answer. 10. y = loga(x2) , then at x = e, is (A) 2e (B) 1/e (C) 2/e (D) none of these Solution: y = loga x2 = = Hence (D) is correct answer. 11. y = , then is (A) (B) (C) (D) none of these Solution: y = + log(cos1x)   1  hence none of these Hence (D) is correct answer. 12. y = , then is (A) (B) (C) (D) Solution: y = sin1 , = = Hence (B) is correct answer. 13. x = , then is (A) (B) (C) (D) none of these Solution: x = log x = tan1  = tan log x = 1 + tan log x y = x2 + x2 tan log x = 2x + 2x tan log x + x sec2 (log x)  = 2x + 2x tan log x + x sec2 (log x) = (x2 + x2 tan log x) + x sec2 (log x) = + 1 + tan2 (log x)  + x = + 2x = + 2x = + 2x Hence (A) is correct answer. 14. x = (log x)tany , then is (A) logx – tany (B) logx + tany (C) (D) Solution: x = (log x)tany  log x = tan y log log x tan y = . Hence (D) is correct answer. 15. If then at x = 0 is (A) 1 (B) 0 (C) –1 (D) none of these Solution: Multiplying numerator and denominator by (1 – x)   So Hence (A) is correct answer. 16. If then f’(x) is equal to (A) sec x(tanx – secx) (B) sec x(secx – tanx) (C) sec x(tanx + secx) (D) none of these Solution: = Thus f’(x) = sec2x + secx tanx = secx (secx + tanx) Hence (C) is correct answer. 17. The differential coefficient of with respect to at x = ½ is (A) – 4 (B) 4 (C) –1 (D) none of these Solution: Let x = cos; then Hence (A) is correct answer. 18. If , then is equal to (A) 1 (B) –1 (C) 0 (D) none of these Solution: Hence (A) is correct answer. 19. If sin(x  y)  exy  2, then is equal to (A) x  0 (B) x  2 (C) x  2 (D) x  3 Solution: If sin(x + y) = ex + y – 2 then, x + y = constant = (say) where '' is the solution of sin = e – 2. Thus, = – 1 20. f(x)  (x2  4) is non-differentiable at (A) x  0 (B) x  2 (C) x  2 (D) x  3 Solution: f(x) = (x2 – 4) |(x – 2) (x – 3)| + cosx f(x) = only points where f(x) may be non–differentiable are x = 2 and x = 3. f(x) = f(2 – 0) = – sin2, f(2 + 0) = – sin2, f(3 – 0) = –5 – sin3 f(3 + 0) = 5 – sin3 Thus, f(x) is differentiable at x = 2 but not at x = 3. 21. If f(x) = logx(logx), then f'(x) at x = e is equal to: (A) e (B) 1/e (C) 2/e (D) 0 Solution : where t = log x  when x = e, t = 1 . Hence (B) is correct answer. 22. f(x)  (A) f(x) is continuous but nondifferentiable (B) f(x) is differentiable (C) f(x) is discontinuous at x = 0 (D) none of these Solution: f(0) = Thus f(x) is differentiable at x = 0 at hence also continuous at x = 0. 23. Exhaustive set of points where f(x)  x is differentiable, is (A) (B) (C) (D) Solution: f(x) = Graph of f(x) indicates that f(x) is differentiable for (–, ). 24. Let , then f(x) is (A) Continuous but nondifferentiable at x  0 (B) Is differentiable at x  0 (C) Discontinuous at (D) none of these Solution: f(0 + 0) = = = = 1 and f(0 – 0) = = = = = –1 as f(0 + 0)  f(0 – 0), thus f(x) is discontinuous at x = 0. FROM AIEEE BOOK 25. If f(x) is a polynomial in x, the second derivative of f(ex) at x = 1 is : (A) ef"(e) + f'(e) (B) (f"(e) + f'(e))e2 (C) e2f"(e) (D) (f"(e)e + f'(e))e Solution. If y = f(ex) (a polynomial in ex), then Hence (D) is correct answer. 26. The differential coefficient of w.r.t. x is : (A) 0 (B) (C) 1 (D) None of these Solution. Hence (C) is correct answer. 27. The differential coefficient of log (tan x) is : (A) 2 sec 2x (B) 2 coesec 2x (C) 2sec2x (D) 2cosec2x Solution. Hence (B) is correct answer. 28. If , then is : (A) (B) (C) (D) Solution. Pur x = tan   y = sin–1(cos 2) Hence (A) is correct answer. 29. Derivative of w.r.t. at is : (A) 2 (B) 4 (C) 1 (D) –2 Solution. Let u = Put x = cos  u = 2, v = sin . Hence (B) is correct answer. 30. If sin(x + y) = log(x + y), then = (A) 2 (B) –2 (C) 1 (D) –1 Solution. Hence (D) is correct answer. 31. If , then value of is : (A) (B) (C) (D) Solution.  y2 = sinx + y Hence (D) is correct answer. 32. If y = log (sin x), then is : (A) –cosec2x (B) sec2x (C) –cosecx cot x (D) secx tanx Solution. y = log (sin x) Hence (A) is correct answer. 33. If , then is : (A) (B) (C) (D) Solution. y = log (1 – x2) – log(1 + x2) Hence (A) is correct answer. 34. If y = a(1 + cos t) and x = a(t – sint), then is : (A) (B) (C) (D) None of these Solution. Hence (C) is correct answer. 35. If , then is : (A) (B) (C) (D) None of these Solution. y = ex+y  log y = x + y Hence (C) is correct answer. 36. If , then at is : (A) (B) 1 (C) –1 (D) 2 Solution. Hence (C) is correct answer. 37. If x = at2, y = 2at, then is : (A) (B) (C) (D) Solution. x = at2 y = 2at [By (1)] Hence (D) is correct answer. 38. Let g(x) be the inverse of the function f(x) and f'(x) . Then g'(x) is : (A) (B) (C) 1 + (g(x))3 (D) 1 + (f(x))3 Solution : Since g(x) is the inverse of f(x)  (gof)(x) = x  g(f(x)) = x  g'(f(x)) f'(x) = 1  g'(y) = 1 + (g(y))2 [ y = f(x)  x = f–1(y) = g(y)]  g'(x) = 1 + (g(x))3. (Replacing y by x) Hence (C) is correct answer. 39. = (A) 0 (B) (C) (D) –1 Solution : Put x = cos    = cos–1 x . Hence (C) is correct answer. 40. If f(x) = loga loga(x), then f'(x) is : (A) (B) (C) (D) Solution : . Hence (A) is correct answer. 41. If yx = xy, then is : (A) (B) (C) (D) Solution : x log y = y log x . Hence (B) is correct ans wer. 42. If pv = 81, then at v = 9 is : (A) 1 (B) –1 (C) 2 (D) None of these Solution : For v = 9, . Hence (B) is correct answer. 43. Differential coefficient of w.r.t. is equal to : (A) 0 (B) –1 (C) 1 (D) None of these Solution : Put and Put . Hence (C) is correct answer. 44. If for x > 0, then f'(x) is : (A) (B) (C) (D) Solution : . Hence (B) is correct answer. 45. If f(x) = cot–1 (cos2x)1/2, then is : (A) (B) (C) (D) Solution : . Hence (C) is correct answer. 46. If , then is : (A) 1 (B) 2 (C) 3 (D) 0 Solution : . Hence (D) is correct answer. 47. If x = a(t + sint), y = a(1 – cost), then is : (A) (B) (C) (D) Solution : . Hence (A) is correct answer. 48. If x2 + y2 = 1, then : (A) yy" – (2y')2 + 1 = 0 (B) yy" + (y')2 + 1 = 0 (C) yy" – (y')2 – 1 = 0 (D) yy" + 2(y')2 + 1 = 0 Solution : . Hence (B) is correct answer. LEVEL II 49. If f'(3) = 5 then is : (A) 5 (B) 1/5 (C) 2 (D) None of these Solution : . Hence (A) is correct answer. 50. If f is twice differentiable function then is : (A) 2f'(a) (B) f''(a) (C) f'(a) (D) f'(a) + f''(a) Solution : . Hence (B) is correct answer. 51. If f(x) = sinx, g(x) = x2, h(x) = logx and F(x) = (hogof)(x) then F''(x) is : (A) 2cosec3x (B) 2cotx2 – 4x2cosec2x2 (C) 2x cotx2 (D) – 2cosec2x Solution : Hence (D) is correct answer. 52. If then the value of y2(0) is : (A) – 1 (B) 0 (C) 1 (D) None of these Solution : . Hence (A) is correct answer. 53. If then (A) 0 (B) 1 (C) – 1 (D) 2 Solution : Differentiating w.r.t. x, Canceling 2y1, Hence (A) is correct answer. 54. If then is : (A) (B) (C) (D) None of these Solution : We are given . . . (1) Also [By (1)] . . . (2) (1) + (2) gives Squaring Canceling 4x, . Hence (C) is correct answer. 55. If then the value of is : (A) n (B) 2n (C) (D) None of these Solution :  Given expression . Hence (B) is correct answer. 56. If y = sin (sinx), and then f(x) is : (A) sin2x sin(cosx) (B) sin2x cos(sinx) (C) cos2xsin(cosx) (D) cos2x sin(sinx) Solution : . Hence (D) is correct answer. 57. If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t then the value of is : (A) 3/2 (B) –5/2 (C) 5/2 (D) –3/2 Solution : . Hence (D) is correct answer. 58. If then is equal to : (A) (B) (C) (D) Solution : Hence (A) is correct answer. 59. If and f'(x) = sin x2 then is: (A) cos x2.f'(x) (B) – cos x2.f'(x) (C) (D) None of these Solution : Let . Hence (C) is correct answer. 60. If y2 = p(x), a polynomial of degree 3 then is equal to: (A) p'''(x) + p'(x) (B) p'''(x) + p''(x) (C) p(x) p'''(x) (D) a constant Solution : . . Hence (C) is correct answer.