Question

Suppose f(π/3) = 3 and f '(π/3) = −5, and let g(x) = f(x) sin(x) and h(x) = cos(x)/f(x). Find the following.

a.g'(π/3)
b. h'(π/3)

Answer 1

`\bf g(x)=f(x)sin(x)\implies \cfrac{dg}{dx}=\cfrac{df}{dx}sin(x)+f(x)cos(x) \\\\\\ g'\left( (\pi )/(3) \right)=f'\left( (\pi )/(3) \right)sin\left( (\pi )/(3) \right)+f\left( (\pi )/(3) \right)cos\left( (\pi )/(3) \right) \\\\\\ g'\left( (\pi )/(3) \right)=-5\cdot \cfrac{√(3)}{2}+3\cdot \cfrac{1}{2}\implies \boxed{g'\left( (\pi )/(3) \right)=\cfrac{3-5√(3)}{2}}\\\\ -------------------------------\\\\`


`\bf h(x)=\cfrac{cos(x)}{f(x)}\implies \cfrac{dh}{dx}=\cfrac{-sin(x)f(x)-cos(x)(df)/(dx)}{\left[ f(x) \right]^2} \\\\\\ h'\left( (\pi )/(3) \right)=\cfrac{-sin\left( (\pi )/(3) \right)f\left( (\pi )/(3) \right)-cos\left( (\pi )/(3) \right)f'\left( (\pi )/(3) \right)}{\left[ f\left( (\pi )/(3) \right) \right]^2}`


`\bf h'\left( (\pi )/(3) \right)=\cfrac{-(√(3))/(2)\cdot 3-(1)/(2)\cdot -5}{3^2}\implies h'\left( (\pi )/(3) \right)=\cfrac{(5-3√(3))/(2)}{9}\implies \boxed{h'\left( (\pi )/(3) \right)=\cfrac{5-3√(3)}{18}}`