# differentiation under integral sign

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:

$V'(D) = -z^*g(I)f(z^*)\frac{\partial z^*}{\partial D} + [-z^*g(I)f(z^*)\frac{\partial z^*}{\partial I} + \int_{z^*}^1 zg'(I)\ dF(z)]\frac{\partial I}{\partial D}$

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 I am a little confused about the relationships between the variables. Does $z$ relate to $z^*$ in any way? Is $I$ independent of $z$ and $z^*$? – mixedmath♦ Jun 7 '11 at 19:15 z* and z are independent. z* is dependent on D and g(I). – adi Jun 7 '11 at 21:15 Does $g(I)$ depend on $D$? What does $I$ mean, i.e. does the integral depend on it, or is it just another parameter? – Gerben Jun 7 '11 at 21:48 I is a parameter. the integral depends on I since it contains a function of I in the integral and the bounds. – adi Jun 8 '11 at 12:22