Answer
$f'(\frac{\pi}{4})=1$
Work Step by Step
Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$
$u(x)=\sin(x) ;u’(x)=\cos(x) $
$v(x)=\sin(x)+\cos(x) ;v’(x)=\cos(x)-\sin(x)$
$f'(x)=(\cos(x))(\sin(x)+\cos(x))+(\sin(x))(\cos(x)-\sin(x))$
$=(2\cos(x)\sin(x))+(\cos^2(x)-\sin^2(x))$
$=\sin(2x)+\cos(2x)$
$f'(\frac{\pi}{4})=sin(\frac{\pi}{2})+\cos(\frac{\pi}{2})=1$.
