# Derivative of a complicated inverse function

$\Phi(\cdot,0,1)$ and $\phi(\cdot,0,1)$ are cdf and pdf of standard normal distribution. $$y=F_\text{mix}(x,\mu,\sigma)=\sum\limits_{i=1}^{K}\lambda_i\Phi\left(\frac{x-\mu_i}{\sigma_i},0,1\right).$$

$x=Q(y)$ is the inverse function of $F_\text{mix}$.

$$\mu_i=\bar{\mu_i}'w,\qquad \sigma_i^2 =w'\Sigma_i w.$$

What is the derivative of $Q(y)$ with respect to $w$?

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I'll assume that by $x=Q(y)$ you mean $x=Q(y,\mu,\sigma)$, where $\mu$ and $\sigma$ are treated as parameters and $F_\text{mix}$ is inverted with respect to $x$. Also, from your expressions for $\mu_i$ and $\sigma_i^2$ I'm guessing that $w$ is a vector, $\Sigma_i$ is a matrix, $\bar{\mu_i}$ is a vector and a prime denotes transposition. (All these things should preferably have been explained in the question.)

This is a great exercise for becoming familiar with partial derivatives and understanding how important it is to keep clear about what's being varied and what's being held fixed. I'll use vertical lines to indicate the variables being held fixed. First,

$$\def\deriv#1#2{\frac{\mathrm d#1}{\mathrm d#2}}\def\pderiv#1#2#3{\left.\frac{\partial#1}{\partial#2}\right|_{#3}}\deriv{Q(y,\mu,\sigma)}w = \pderiv x\mu{y,\sigma} \deriv\mu w+\pderiv x\sigma{y,\mu}\deriv\sigma w\;.$$

To find the derivatives of $x$, differentiate the defining equation for $y$:

$$\begin{eqnarray} 0 &=& \pderiv y\mu{y,\sigma} \\ &=& \pderiv yx{\mu,\sigma}\pderiv x\mu{y,\sigma}+\pderiv y\mu{x,\sigma}\pderiv \mu\mu{y,\sigma}+\pderiv y\sigma{x,\mu}\pderiv \sigma\mu{y,\sigma} \\ &=& \pderiv yx{\mu,\sigma}\pderiv x\mu{y,\sigma}+1\cdot\pderiv y\mu{x,\sigma}+0\cdot\pderiv y\sigma{x,\mu} \\ &=& \pderiv yx{\mu,\sigma}\pderiv x\mu{y,\sigma}+\pderiv y\mu{x,\sigma}\;, \end{eqnarray}$$

and thus

$$\pderiv x\mu{y,\sigma}=-\pderiv y\mu{x,\sigma}\left(\pderiv yx{\mu,\sigma}\right)^{-1}\;,$$ and likewise for $\partial x/\partial\sigma$. Thus we have

$$\deriv{Q(y,\mu,\sigma)}w =-\left(\pderiv yx{\mu,\sigma}\right)^{-1}\left( \pderiv y\mu{x,\sigma} \deriv\mu w+\pderiv y\sigma{x,\mu}\deriv\sigma w\right)\;.$$

Perhaps you can take it from there, since these are all derivatives of the functions given explicitly; let me know if you need further assistance.

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