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As the title says, I am trying to find the derivative of the inverse cumulative distribution function for the standard normal distribution. I have this figured out for one particular case, but there is an extra layer of complexity that has be stumped.

Let $0 \le p \le 1$ and let $z = \Phi^{-1}(p)$, where $\Phi^{-1}(p)$ is the inverse cumulative distribution function for the standard normal distribution. Then: $$\frac{\partial \Phi^{-1}(p)}{\partial p} = \left(\frac{\partial \Phi(z)}{\partial z}\right)^{-1},$$ where $\Phi(z)$ is the cumulative distribution function for the standard normal distribution. This yields: $$= \left(\frac{1}{\sqrt{2\pi}} \exp(-z^2/2) \right)^{-1} = \frac{\sqrt{2\pi}}{\exp(-z^2/2)}.$$

I think/hope this is right so far.

But now I have $p_1$ and $p_2$ and I need to find the derivative of $$\frac{\partial \Phi^{-1}\left(\frac{p_1}{p_1+p_2}\right)}{\partial p_1}$$ and $$\frac{\partial \Phi^{-1}\left(\frac{p_1}{p_1+p_2}\right)}{\partial p_2}.$$ Any help would be appreciated.

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$\Phi:\mathbb R \to (0,1)$ and $\Phi^{-1}:(0,1) \to \mathbb R$ are strictly increasing continuous bijective functions

If $z=\Phi^{-1}(p)$ then $p=\Phi(z)$ and $\dfrac{dp}{dz}=\Phi^\prime(z)=\phi(z)$, so $\dfrac{dz}{dp} = \dfrac{1}{\phi(z)}= \dfrac{1}{\phi(\Phi^{-1}(p))}$

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This can be done using chain rule, note that

$$ f(f^{-1}(x)) = x $$ and therefore $$ \frac{d}{dx}f(f^{-1}(x)) = \frac{d}{dx}x = 1 \tag 1 $$ and also by chain rule $$ \frac{d}{dx}f(f^{-1}(x)) = f'(f^{-1}(x)) \cdot (f^{-1})'(x) \tag 2 $$

So by $(1)$ and $(2)$ we have $$ \begin{align} f'(f^{-1}(x)) \cdot (f^{-1})'(x) &= 1 \\ (f^{-1})'(x) &= \frac{1}{f'(f^{-1}(x))} \end{align} $$ So in your case, with $f = \Phi$ we have $(\Phi^{-1})'(x) = \frac{1}{\Phi'(\Phi^{-1}(x))}$

where for a standard normal distribution

$$ \Phi^{-1}(x) = \sqrt{2} \operatorname{erf}^{-1}(2x-1) $$ and $\Phi'(x)$ is just the pdf of a standard normal, i.e. $$ \Phi'(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} $$

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