How do I find g'(x) at a certain x value given f'(x)?

Question

so, I'm supposed to evaluate $g'(\frac{\pi}3)$ if $g(x) = f(sec (x))$, given the following conditions:

$f'(x) = 6x *f(x)$ and $f(2) = 2$.

I've tried solving for $g'(\frac{\pi}3)$ by substituting $sec(x)$ to $f'(x)$,

$f'(sec (x)) = 6(sec(x))*sec(x)) = 6sec^2(x) = 6sec^2(\frac{\pi}3)$ = 24

however, evaluating like this seems to be wrong.

I think the way to solve this has something to do with the chain rule? But, currently, I'm at a loss on how to go about this problem. Can someone help?

Thank you in advance!

Answer 1

$$ g(x) = f( \sec x)) \implies g'(x) = f'( \sec x)) \sec(x)\tan(x)$$

At $x= \pi/3$ we get $$ g'(\pi/3) = f'( 2) (2)(\sqrt 3) = 4\sqrt 3$$

Answer 2

Hint:

$g(x) = f(sec (x))$

Chain rule says $$(f\circ g)'=(f'\circ g)g'$$

By chain rule $g'(x)=f'(\sec(x))(\sec(x)\tan(x))$

We are using the fact that derivative of $\sec(x)$ is $\sec(x)\tan(x)$