One Variabe Chain Rule

Question

$f(x)$ and $g(x)$ differentiable functions. Let $h(x)=f(g(x))$. It is given that $g(2)=1$, $g'(2)=e^{-2}$ and $h'(2)=2/e$. I need to find $f'(1)$.

I know how to find $f'(2)$. But $f'(1)$? $f'(2)$ should be $2e$.

Answer 1

Hint Note that for any $x$, we have $$ h'(x) = f'(g(x))g'(x) $$ now, let $x = 2$.

Answer 2

$h'(x) = f'(g(x))g'(x) \implies h'(2) = f'(g(2))g'(2)$. But $g(2) = 1$, then:

$h'(2) = f'(g(2))g'(2) \implies h'(2) = f'(1)g'(2) \implies f'(1) = \dfrac{h'(2)}{g'(2)} = \dfrac{2}{e}\dfrac{1}{e^ {-2}} = 2e.$

So, $f'(1) = 2e$.