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