Derivative of the Inverse Cumulative Distribution Function for the Standard Normal Distribution 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.

$\Phi:\mathbb R \to (0,1)$ 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 $\dfrac{dz}{dp} = \dfrac{1}{\phi(z)}= \dfrac{1}{\phi(\Phi^{-1}(p))}$
