If $y = x^{sin x}$, $x > 0$, find $\dfrac{dy}{dx}$

The correct answer is $x^{sin x}$ $\left[ \dfrac{sinx}{x} + logx cosx \right] $

$y = x^{sin x}$

Taking log on both sides,

$log$ $y$ = $log$ $ x^{sin x}$


As $log$ $p^{n} = n$ $log p$, the above can be written as

$log$ $y$ = $sinx$ $log x$


Differentiating both sides with respect to $x$ and using the product rule,

$\dfrac{1}{y}$ $\dfrac{dy}{dx}$ = $\dfrac{sinx}{x}$ + $logx$ $cosx$


$\dfrac{dy}{dx}$ = $y$ $\left[ \dfrac{sinx}{x} + logx cosx \right] $


$\dfrac{dy}{dx}$ = $x^{sin x}$ $\left[ \dfrac{sinx}{x} + logx cosx \right] $