## Calculus: Early Transcendentals 8th Edition

(a) $$g'(\pi/3)=-\sqrt 3+2$$ (b) $$h'(\pi/3)=\frac{-2\sqrt3+1}{16}$$
$f(\pi/3)=4$, $f'(\pi/3)=-2$, $g(x)=f(x)\sin x$ and $h(x)=\frac{\cos x}{f(x)}$ (a) First, find $g'(x)$ $$g'(x)=[f(x)\sin x]'$$ Apply Product Rule: $$g'(x)=f'(x)\sin x+f(x)(\sin x)'$$ $$g'(x)=f'(x)\sin x+f(x)\cos x$$ So, $$g'(\pi/3)=f'(\pi/3)\sin(\pi/3)+f(\pi/3)\cos(\pi/3)$$ $$g'(\pi/3)=-2\times\frac{\sqrt 3}{2}+4\times\frac{1}{2}$$ $$g'(\pi/3)=-\sqrt 3+2$$ (b) First, find $h'(x)$ $$h'(x)=[\frac{\cos x}{f(x)}]'$$ Apply Quotient Rule: $$h'(x)=\frac{(\cos x)'f(x)-\cos x f'(x)}{f^2(x)}$$ $$h'(x)=\frac{-\sin x f(x)-\cos x f'(x)}{f^2(x)}$$ So, $$h'(\pi/3)=\frac{-\sin(\pi/3)f(\pi/3)-\cos(\pi/3)f'(\pi/3)}{f^2(\pi/3)}$$ $$h'(\pi/3)=\frac{-\frac{\sqrt 3}{2}\times4-\frac{1}{2}\times(-2)}{4^2}$$ $$h'(\pi/3)=\frac{-2\sqrt3+1}{16}$$