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