Given $F(z)$ is a cumulative distribution function
$z^* = \frac{D}{g(I)}$ and
$V(D) = \int_{z^*}^1 zg(I)\ dF(z)$
How can you differentiate $V(D)$ with respect to D?
I assume that it has to do with differentiation under the integral sign, but I am not sure how to perform it when the integral is over $dF(z)$ .
Edit: It seems that the following solution is correct, but I still can't seem to understand the steps to get it:
$V'(D) = -z^*g(I)f(z^*)\frac{\partial z^*}{\partial D} + [-z^*g(I)f(z^*)\frac{\partial z^*}{\partial I} + \int_{z^*}^1 zg'(I)\ dF(z)]\frac{\partial I}{\partial D}$
