Question

If y= log( xe^x), then derivative is given by

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = log(x {e}^{x} )$

$\rm :\longmapsto\:y = logx + log {e}^{x}$

$\: \: \: \: \: \: \: \: \green{\boxed{ \bf{ \: \because \: logxy = logx \: + \: logy }}}$

$\rm :\longmapsto\:y = logx \: + x \: loge$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf{ \: \because \: log {x}^{y} = y \: logx }}}$

$\rm :\longmapsto\:y = logx + x \times 1$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf{ \: \because \: loge = 1 }}}$

$\rm :\longmapsto\:y = logx + x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx}( logx + x)$

$\rm :\longmapsto\:\dfrac{dy}{dx} =\dfrac{d}{dx} logx +\dfrac{d}{dx} x$

$\bf :\longmapsto\:\dfrac{dy}{dx} =\dfrac{1}{x} + 1$

Additional Information :-

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}k = 0}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}x = 1}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}logx = \dfrac{1}{x} }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} {e}^{x} = {e}^{x} }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} {a}^{x} = {a}^{x}logx }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} sinx = cosx }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} cosx = - sinx }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} cotx = - {cosec}^{2} x }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} tanx = {sec}^{2} x }}}$