$\frac{d\Phi^{-1}(y)}{dy} = \frac{1}{\frac{d}{dy}[\Phi(\Phi^{-1}(y))]}$?

Question

If $\Phi(y)$ is a monotonic decreasing function is true that

$$\frac{d\Phi^{-1}(y)}{dy} = \frac{1}{\Phi'(\Phi^{-1}(y))}$$

If so, how?

It works for $y = \Phi(x) = e^{-x}, \quad \Phi^{-1}(y) = -log(y), \quad \frac{d\Phi^{-1}(y)}{dy} = \frac{-1}{y}, \quad $

Answer 1

use the fact $$ \Phi\left(\Phi^{-1}(y)\right) = y$$ and the chain rule for the derivative.

Answer 2

The fact that $${\phi'(y)} = \frac{1}{f'(x)}$$ is a well known theorem for all invertible functions where $\phi(y) = f^{-1}(y)$. More on it here.