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Mathematics Session Differentiation - 3 Session Objectives Logarithmic Differentiation, Infinite Series Differentiation of Parametric Functions, Functions w. r. t. Another Function Second Order Derivatives Class Exercise Logarithmic Differentiation (1) When the functions of the form f x g x we first take logarithms and then differentiate. (2) When the function is a product of many simpler functions. Sin2x Sin3x Sin4x In this case, logarithm converts the product into a sum. Example-1 x y=x x Differentiate: w.r.t. x. Solution: x y=x x Taking log of both sides, we get x = xxlogx logy = logx x Taking log again, we get loglogy =log(xxlogx) loglogy = xlogx +loglogx Solution Cont. 1 1 dy 1 1 1 × = x× +logx×1+ × logy y dx x logx x dy 1 = ylogy 1+logx + dx xlogx x Substituting, y = x and logy = xxlogx x xx x dy 1 =x x logx 1+logx + dx xlogx xx x dy 1 2 =x x logx + logx + dx x Example-2 Differentiate: Sin2x Sin3x sin4x w.r.t.x. Solution: Let y = Sin2x Sin3x sin4x Taking log of both sides, we get logy = logsin2x +logsin3x +logsin4x 1 dy 1 1 1 × = cos2x×2+ cos3x×3+ cos4x×4 y dx sin2x sin3x sin4x dy = y 2cot2x+3cot3x+4cot4x dx dy = sin2xsin3xsin4x 2cot2x+3cot3x+4cot4x dx Example-3 dy If x + y = a , find . dx y x b Solution: We have xy + yx = ab Let u = xy and v = yx u+ v = ab du dv + =0 dx dx Con. Now u = xy logu = ylogx Taking log on both sides dy 1 du 1 = y +logx× u dx x dx dy du y = u +logx dx dx x dy du = xy-1 y + xlogx dx dx Differentiating both sides Con. Also, v = yx logv = xlogy 1 dv 1 dy x log y v dx y dx Taking log on both sides Differentiating dy x + ylogy dx dv =v dx y dv x-1 dy =y x + ylogy dx dx both sides Con. du dv + =0 dx dx dy x-1 dy y + xlogx dx + y x dx + ylogy = 0 y-1 x y x-1 x logx + xy dy + yxy-1 + yxlogy = 0 dx y-1 dy + yxlogy yx =- y x-1 dx x logx + xy Differentiation of Infinite Series y=x ... xx y = xy Taking log on both sides, we get logy = ylogx 1 dy 1 dy = y× +logx y dx x dx y dy 1 logx = dx y x dy y2 = dx x 1- ylogx Parametric Functions When two variables x and y are expressed in the form x = f(t), y = g(t) , where t is a parameter dx dy = f' t , = g' t dt dt dy dy = dt dx dx dt Example-4 If x = acos3t and y = asin3t, find Solution: dy dx We have x = acos3t and y = asin3t Differentiating w.r.t. t, we get dx dy =-3acos2t sint and =3asin2t cost dt dt dy 2 dy 3asin tcost = dt = = -tant 2 dx dx -3acos tsint dt Differentiation of Functions w. r. t. Another Function u = f(x) and v = g(x) Derivative of f(x) with respect to g(x) du du dx = is dv dv dx Example-5 Differentiate tan2 x w.r.t. cos2 x. Solution: Let y = tan2x and z = cos2x dy d = tan2 x dx dx 2 tan x sec 2 x dz d Also = cos2 x dx dx = 2cosx -sinx Con. dy dy dx = × dz dx dz 1 = 2tanxsec x× -2sinxcosx 2 -sinx 1 = × 3 cos x sinxcosx = -sec4 x Example-6 -1 Differentiate sin 2x 1 + x2 -1 w.r.t. cos 1 - x2 1 + x2 Solution: 2 -1 1 - x and v = cos 2 2 1+ x 1+ x -1 Let u = sin 2x Putting x = tanq 2 2tanθ -1 1 - tan θ and v = cos 2 2 1+ tan θ 1+ tan θ -1 u = sin u = sin-1 sin2θ and v = cos-1 cos2θ , if 0 < x < 1. Solution Cont. u = 2θ and v = 2θ Differentiating w.r.t. x, we get du dv = 2 and =2 dθ dθ du du dθ 2 = = =1 du 2 dv dθ Second Order Derivatives y = ƒ x dy First order differential coefficient dx d dy d2 y = Second order differential coefficient 2 dx dx dx Example-7 x2 dy If y = log , find . e2 dx Solution: x2 y = log e2 y = logx2 - loge2 y = 2logx - 2 dy 2 = -0 dx x d dy 2 d2y 2 = = dx dx x2 dx2 x2 Example-8 If y = ex sinx + cosx , prove that d2 y dx2 -2 dy + 2y = 0. dx Solution: We have y = ex sinx +cosx dy = ex cosx - sinx + ex sinx + cosx dx dy = 2ex cosx dx d2 y dx2 = 2ex -sinx + 2ex cosx Con. d2 y dx2 LHS = = 2ex cosx - sinx d2 y dx2 -2 dy + 2y dx = 2ex cosx - sinx - 2×2excosx +2×ex sinx +cosx = 0 = RHS Thank you