I'm having a really hard time with this question I keep googling for advice but can't find anything solid that's similar! Please help. I'm not sure if I should derive first or find the inverse first? And also: I am not sure what to do with the $f(0)=2$ information.
calculate $$(f^{-1})'(2)$$ while $$f(0) = 2$$
$$f(x)=4x^3+2 \sin x+2 \cos x$$
I'm really not sure how to go about answering this. Thank you!
Use the fact that, by the chain rule $$ 1 = \frac{d}{dx}x = \frac{d}{dx}f(f^{-1}(x)) = f'(f^{-1}(x))\left( \frac{d}{dx} f^{-1}(x)\right) $$ to conclude that $$ \frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}. $$ You are asked to find $(f^{-1})'(2)$ which you can now see is just $\frac{1}{f'(f^{-1}(2))}$. So you should find $f'$ and apply it to $f^{-1}(2)$, which you are told is $0$: $$ (f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(0)} = (\left.12x^2 + 2\cos x - 2\sin x \right|_{x=0})^{-1} = 1/2. $$