Given $f(x) = x^3+ 3\sin x + 2\cos x$ and $a = 2$
What is $({f^-}^1)'(a)$?
This is my understanding, I'm figuring out how to continue:
let $y=({f^-}^1)(x)$
$$f(y)=x$$
$$f'(y) \frac{dy}{dx} = 1$$
$$\frac{dy}{dx} = \frac{1}{f'({f^-}^1)(x)}$$
$$({f^-}^1)'(a) = \frac{1}{f'({f^-}^1)(a)}$$
$$f(x) = x^3 +3\sin x + 2\cos x$$
let $a = f(x)$ => $x={f^-}^1(a)$
$$x^3 +3\sin x + 2\cos x-2=0$$
Let me know if I'm heading in the right direction. I used the same method to solve the same type of question but in polynomial form only.