How can I calculate the derivative of the standard normal inverse. I think the derivative of $\Phi^{-1}(x)$ is $$\frac{1}{\phi(\Phi^{-1}(x))}.$$ I would like to know how to find the derivative of $$\Phi^{-1} \left(\frac{x}{c}\right),$$ where $c$ is a known constant.
Any help would be much appreciated.
On
It's done just as with any inverse function, using the same method by which one proves that $$\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}\text{ or that }\dfrac{d}{dx}\log_e x = \dfrac1x. $$
Suppose $y=\Phi^{-1}(x)$.
Then $\Phi(y) = x$.
So $\Phi'(y) = \dfrac{dx}{dy}$, or $\varphi(y) = \dfrac{dx}{dy}$.
Therefore $$ \frac{dy}{dx} = \frac{1}{\varphi(y)} = \frac{1}{\varphi(\Phi^{-1}(x))}. $$
It may be questioned whether treating $\dfrac{dy}{dx}$ in this way, as if $dy$ and $dx$ are actual numbers, is permissible. But in fact it just amounts to an application of the chain rule.
Apply the chain rule:
$$\frac{d}{dx}\left(\Phi^{-1}\left(\frac{x}{c}\right)\right)=\frac{d}{dx}\Phi^{-1}{\bigg|}_{\frac{x}{c}}\cdot\frac{d}{dx}\left(\frac{x}{c}\right)=\frac{1}{\Phi'(\Phi^{-1}(\frac{x}{c}))}\cdot\frac{1}{c}$$