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