How Do I Compute the Derivative of an Integral with a Single Variable Change?

[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!

 

[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

 

[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.