# Thread: Derivatives of Inverse Functions

1. ## Derivatives of Inverse Functions

I am absolutely stumped on this question for calculus, but I want to try and see if I can get it solved or partially solved.

Suppose f is invertible and differentiable with f'(x) not equal to zero. Use implicit differentiation of the relation f(f-1 (x)) = x to show that f-1 is differentiable with (f-1)′ (x) = 1 / f′(f-1(x)).

I have looked at some examples in the book and class notes but have no idea how to get started to start solving this problem. If anyone can help me out or give me a start on how to solve the problem that would be greatly appreciated.

2. So start with $\displaystyle f \left( f^{-1}[x] \right) = x$ then differentiate this expression -> $\displaystyle \frac{d}{dx} f \left( f^{-1}[x] \right) = 1$. To differentiate the left hand side, use the chain rule. So:

$\displaystyle \frac{d}{dx} f \left( f^{-1}[x] \right) = f' \left( f^{-1}[x] \right)*[f^{-1}(x)]^{'}$.

So combine two expressions to get: $\displaystyle f' \left( f^{-1}[x] \right)*[f^{-1}(x)]^{'}=1$

Now there is one step left which is to solve for what you are looking for.