Find $g'(\pi)$ where $g(t) = f(F(t))$

Question

If $f(x,y,z) = \sin x$ and $F(t) = (\cos t, \sin t, t)$, find $g'(\pi)$ where $g(t) = f(F(t))$.

First, I compute $$\begin{align} g(t) &= f(F(t)) \\ &=f(\cos t, \sin t, t) \\ &= \sin(\cos t) \end{align}$$

Then I find $g'(t) = \cos(\cos t) \cdot (-\sin t)$, but I think I am doing something wrong here and could use some help being pointed in the right direction. Thanks

Answer 1

There is nothing wrong from what you did and it follows from it that $g'(\pi)=0$. Of course, you could also have applied the chain rule to $f$ and $F$, but you don't have to do that.