Chapter 5
dy y dx = sin (log ) log (sin )d dx x x xdx dx+ or 1 dy y dx = 1(sin ) log cosx x xx + or dy dx = sin cos logxy x xx + = sin sin cos logx xx x xx + = sin 1 sinsin cos logx xx x x x x− ⋅ + ⋅ Example 30 Find dy dx , if yx + xy + xx = ab. Solution Given that yx + xy + xx = ab. Putting u = yx, v = xy and w = xx, we get u + v + w = ab Therefore 0du dv dw dx dx dx+ + = ... (1) Now, u = yx. Taking logarithm on both sides, we have log u = x log y Differentiating both sides w.r.t. x, we have 1 du u dx⋅ = (log ) log ( )d dx y y xdx dx+ = 1 log 1dyx yy dx⋅ + ⋅ So du dx = log logxx dy x dyu y y yy dx y dx + = + ... (2) Also v = xy Taking logarithm on both sides, we have log v = y log x Differentiating both sides w.r.t. x, we have 1 dv v dx⋅ = (log ) logd dyy x xdx dx+ = 1 log dyy xx dx⋅ + ⋅ So dv dx = logy dyv xx dx + = logy y dyx x x dx + ... (3) Again w = xx Taking logarithm on both sides, we have log w = x log x. Differentiating both sides w.r.t. x, we have 1 dw w dx⋅ = (log ) log ( )d dx x x xdx dx+ ⋅ = 1 log 1x xx⋅ + ⋅ i.e. dw dx = w (1 + log x) = xx (1 + log x) ... (4) From (1), (2), (3), (4), we have log logx yx dy y dyy y x xy dx x dx + + + + xx (1 + log x) = 0 or (x . yx - 1 + xy . log x) dy dx = - xx (1 + log x) - y . xy-1 - yx log y Therefore dy dx = 1/1 [ log . (1 log )] . log
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− / − / − + + + / + MATHEMATICS134
sin (tan -1 e-x) 5. log (cos ex) 6. 2 5 ...x x xe e e+ + + 7. , 0xe x > 8. log (log x), x > 1 9. cos , 0log x xx > 10. cos (log x + ex), x > 0 5.5. Logarithmic Differentiation In this section, we will learn to differentiate certain special class of functions given in the form y = f (x) = [ u(x)]v (x) By taking logarithm (to base e) the above may be rewritten as log y = v(x) log [ u(x)] Using chain rule we may differentiate this to get 1 1( ) ( ) dy v xy dx u x⋅ = ⋅ . u′(x) + v′(x) . log [u(x)] which implies that ( ) ( ) log ( )( )
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= ⋅ ′ + ′ ⋅ The main point to be noted in this method is that f (x) and u(x) must always be positive as otherwise their logarithms are not defined. This process of differentiation is known as logarithms differentiation and is illustrated by the following examples: Example 27 Dif ferentiate 2/2 ( 3) ( 4) / 3 4 5 / x x / x x − + + + w.r.t. x. Solution Let 2/2 ( 3) ( 4) (3 4 5) x xy x x − += + + Taking logarithm on both sides, we have log y = 1 2 [log (x - 3) + log (x2 + 4) - log (3x2 + 4x + 5)] Now, differentiating both sides w.r.t. x, we get 1 dy y dx⋅ = 2 2 1 1 2 6 4 2 ( 3) 4 3 4 5 x x
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+ + − − + + + or dy dx = 2 2 1 2 6 4 2 ( 3) 4 3 4 5
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+ + − − + + + = 2 2 2 1 ( 3)( 4) 1 2 6 4 2 ( 3)3 4 5 4 3 4 5
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− + + + − −+ + + + + Example 28 Differentiate ax w.r.t. x, where a is a positive constant. Solution Let y = ax. Then log y = x log a Differentiating both sides w.r.t. x, we have 1 dy y dx = log a or dy dx = y log a Thus ( )xd adx = ax log a Alternatively ( )xd adx = log log( ) ( log )x a x ad de e x adx dx= = ex log a . log a = ax log a. MATHEMATICS132 Example 29 Differentiate xsin x, x > 0 w.r.t. x. Solution Let y = xsin x. Taking logarithm on both sides, we have log y = sin x log x Therefore