## Calculus 8th Edition

$y'=\frac{x(cotx)}{ln4}+log_{4}(sinx)$
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, $y'=\frac{x(cotx)}{ln4}+log_{4}(sinx)$