## Calculus 10th Edition

$f'(x)=\frac{xcos(x)-sin(x)}{x^2}$ $f'(\frac{\pi}{6})=\frac{3\pi\sqrt 3 -18}{\pi^2}$
Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=sin(x); u'(x)=cos(x)$ $v(x)=x; v'(x)=1$ $f'(x)=\frac{(cos(x))(x)-(sin(x)(1)}{x^2}=\frac{xcos(x)-sin(x)}{x^2}$ $f'(\frac{\pi}{6})=\frac{(\frac{\pi}{6})(cos(\frac{\pi}{6})-sin(\frac{\pi}{6})}{(\frac{\pi}{6})^2}=\frac{3\pi\sqrt 3 -18}{\pi^2}$