What's equivalent to the following expression?

$\displaystyle \frac { \partial} {\partial p}\int _p^{p_0} f(\alpha,\beta)d\alpha $

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- Sep 20th 2009, 10:29 AMaltave86Derivative of a definite integral
What's equivalent to the following expression?

$\displaystyle \frac { \partial} {\partial p}\int _p^{p_0} f(\alpha,\beta)d\alpha $ - Sep 20th 2009, 10:10 PMredsoxfan325
Assume that $\displaystyle \int f(\alpha,\beta)\,d\alpha = F(\alpha,\beta)$. Thus,

$\displaystyle \int_p^{p_0}f(\alpha,\beta)\,d\alpha = F(\alpha,\beta)\big|_p^{p_0}=F(p_0,\beta)-F(p,\beta)$

Note that all operations you are doing are either w.r.t. $\displaystyle \alpha$ or $\displaystyle p$, so $\displaystyle \beta$ and $\displaystyle p_0$ should be treated as constants, and therefore $\displaystyle F(p_0,\beta)$ is a constant. Call it $\displaystyle C$.

Now we want $\displaystyle \frac{\partial}{\partial p}[C-F(p,\beta)]=\frac{\partial}{\partial p}[-F(p,\beta)]$ because the derivative of a constant is zero.

Remember how we defined $\displaystyle F$ though. We defined it so that $\displaystyle F~'=f$. So

$\displaystyle \frac{\partial}{\partial p}[-F(p,\beta)]=\boxed{-f(p,\beta)}$