# Derivative of an integral

1. Feb 11, 2012

### carlosmg1982

Hi all,

This is my question. Suppose that I have a continuum of people on the measure [0,1]. Then, I want to aggregate their spending as follows,

$\int^{1}_{0}(s^{\alpha}(i))di$

where i is any person in [0,1]. Suppose that I want to compute the derivative of the previous expression if $s^{\alpha }$ goes up for only one specific i$\in[0,1]$. That is, I want,

$\frac{d(\int^{1}_{0}(s^{\alpha}(i))di)}{d(s(i))}$

How can I compute that derivative...? First, I thought that it would be zero since the contribution to the integral is infinitesimal, but I am not sure about that...

Thank you!

Last edited: Feb 11, 2012
2. Feb 11, 2012

### HallsofIvy

Is $s^{\alpha}$ s to the $\alpha$ power? If so Leibniz's rule:
$$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t)dt= \frac{d\beta}{dx}f(x,\beta(x))- \frac{d\alpha}{dx}f(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial f}{\partial x}dt$$
Works: the derivative is
$$\alpha \int_0^1 s^{\alpha- 1}(i)di$$

3. Feb 11, 2012

### carlosmg1982

Sorry, I did not write the problem properly. There is one specific J$\in(0,1)$, and I want
$\frac{d(\int^{1}_{0}(s(i)^{\alpha})di)}{d(s(J))}$. That is, the derivative with respect to that specific s(J). And yes, it is to the power of $\alpha$.