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$$\frac{d}{du}\int_{g(0)}^{g(u)}f(u,t)\, dt$$

I have an integral of a function $f$ which has variables $u$ and $t$. Ordinarily I could just move the $d/du$ inside since the integral is with respect to $t$. But the bound on the integral is a function of $u$.

For what it's worth, $g$ is a linear function.

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Yes, use the Leibniz Rule:

\begin{align} {d\over du}\int_{g(0)}^{g(u)} f(t,u)\,dt&=\int_{g(0)}^{g(u)} {\partial f\over \partial u}(g(u),u)\,dt+f(g(u),u)g'(u)-f(g(0),u){d\over du}[g(0)]\\ &=\int_{g(0)}^{g(u)} {\partial f\over \partial u}(g(u),u)\,dt+f(g(u),u)g'(u). \end{align}

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Oh good. For some reason I've never seen that one before! –  Forever Mozart Jan 29 '13 at 18:26
    
First time for everything! ;-) –  JohnD Jan 29 '13 at 18:27
    
JohnD, to be more specific I am trying to find is $\frac{d}{du}$ at u=0 $\int_{t1}^{t0+u}F(f(u,t),g(u,t))\, dt$ Could you help me with this one? –  Forever Mozart Jan 29 '13 at 19:54

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