Question

Suppose that f(pi/3)=4 and f'(pi/3)=-2, and let g(x)= (cosx)/f(x). find g'(x) (pi/3)

Answer 1

`g(x)=(\cos x)/(f(x))`

`\implies g'(x)=-(f(x)\sin x+f'(x)\cos x)/(f(x)^2)`

You know that
`f\left(\frac\pi3\right)=4` and
`f'\left(\frac\pi3\right)=-2`, so


`g'\left(\frac\pi3\right)=-(4\sin\frac\pi3-2\cos\frac\pi3)/(4^2)=(1-2\sqrt3)/(16)`