Answer
$y'=x^{sinx}(\frac{sinx}{x}+lnx cosx)$
Work Step by Step
Given: $y =x^{sinx}$
Taking logarithmic on both sides of the function
$y =x^{sin x}$
$lny=sinx .lnx$
Take implicit differentiation with respect to $x$.
Apply product rule of differentiation.
$\frac{d}{dx}(lny)=\frac{d}{dx}(sinx. lnx)$
$\frac{1}{y}\frac{d}{dx}(y)=sinx\frac{d}{dx}(lnx)+lnx\frac{d}{dx}(sinx)$
$\frac{d}{dx}(y)=y[sin\times(\frac{1}{x})+lnx(cosx)]$
Hence, $y'=x^{sinx}(\frac{sinx}{x}+lnx cosx)$
