# Differentiating inside an integral sign

I'm reading John Taylor's Classical Mechanics book and I'm at the part where he's deriving the Euler-Lagrange equation.

Here is the part of the derivation that I didn't follow:

I don't get how he goes from 6.9 to 6.10 by partial-differentiating the term inside the integral. If this is allowed, I was probably missed my calculus class the day it was covered. Can someone tell me more about this? Which part of calculus is this from?

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Be sure to check out the 'in popular culture' section of the wikipedia entry on differentiating under the integral sign for the anecdote about Feynman. Apparently, it is a long-standing problem that this powerful tool is not typically taught in introductory calculus classes. – kleingordon Mar 10 '12 at 5:23

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@kleingordon interesting. It was never taught to us either, but our teacher used it to solve a sum in a beautiful manner. Fortunately, he gave a quick intuitive explanation when he realised that none of us knew it. I guess the rule is so obvious to some that they forget to teach it. – Manishearth♦ Mar 10 '12 at 5:32

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@Manishearth, I agree that it seems like a pretty obvious thing to do (at least in the simple case when the limits aren't functions of your variable), but often enough in math doing something that seems obvious can get you into trouble, especially when you're relatively inexperienced. I think there should be a major effort to put this into the standard calc curriculum. – kleingordon Mar 10 '12 at 5:37
Well, now I know for certain that I'm not going to be a Feynmann. :( – Joebevo Mar 10 '12 at 5:39
Huh, interesting, I never realized that this was a common misunderstanding. Good question Joebevo :-) – David Zaslavsky♦ Mar 11 '12 at 6:20
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## 1 Answer

It's known as the Leibniz integral rule. As long as $\alpha$ is not the variable being integrated over, then $$\frac{\mathrm{d}}{\mathrm{d}\alpha}\int f(x,\alpha) \mathrm{d}x=\int\frac{\partial f(x,\alpha)}{\partial \alpha}\mathrm{d}x$$

$x$ will not be present outside the integral anyways (due to limits of the integral). As it is, while differentiating wrt $\alpha$, $x$ is constant. So it becomes a partial derivative inside.

You may want to check out the proof and more complicated forms (involving limits as functions of $\alpha$) on the linked wiki page.

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It's not that intuitive to me, but I can see that it works by plugging in simple functions. Thanks. – Joebevo Mar 10 '12 at 6:06

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Don't worry, I hated it too at first :) – Manishearth♦ Mar 10 '12 at 6:24
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