Answer
$y'=\frac{x(cotx)}{ln4}+log_{4}(sinx)$
Work Step by Step
Differentiate y with respect to x.
$y'=\frac{1}{ln4}\frac{d}{dx}(xlog(sinx))$
$=\frac{1}{ln4}[x\frac{d}{dx}(logsinx)]+log_{4}(sinx)\frac{d}{dx}(x)$
$=\frac{1}{ln4}[\frac{x}{sinx}(cosx)]+log_{4}(sinx)$
Hence, $y'=\frac{x(cotx)}{ln4}+log_{4}(sinx)$