# Partial Derivative of an integral, how do you do this?

1. Apr 26, 2007

### hanson

Hi all.
How to do the partial differentiation with this integral? (please see the attachment)
I find no place to start tackling this problem...

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2. Apr 26, 2007

### hanson

Can I just put the partial derivative into the integral?

3. Apr 26, 2007

### AlphaNumeric

Assuming everything is 'nice' then yes you can. There's probably a pathological counter example to it being generally true but for most things you can just put the derivative under the integral.

4. Apr 26, 2007

### ObsessiveMathsFreak

Whenever you take the derivative of an integral, be it partial or otherwise, you must use Leibniz's Rule for Integration.

Now, sometimes authors will use a partial derivative outside the integral sign to mean that they're just going to take that partial derivative inside the integral, and use a total to mean that they will use the full Liebnitz rule. However, I'm not convinced that these authors are correct in this, however what they mean to say may be correct, and they're just not saying it properly.

Long story short, always use Leibnitz's rule when differentiating an integral unless you have a very, very, very good reason(from the "physics" of the problem) for thinking it should be otherwise.

5. Apr 26, 2007

### HallsofIvy

Staff Emeritus
You don't say anything about limits of integration so I assume that is an "indefinite integral" (if the limits of integration involve t, that would change the result. If they did NOT involve x, this would not be a partial derivative).

$$\frac{\partial}{\partial t}\int \eta \eta_{xxx}dx=\int\frac{\partial \eta \eta_{xxx}}{\partial t}dx$$
Now use the product rule
$$= \int \left(\frac{\partial \eta}{\partial t}\eta_{xxx}+\eta \frac{\partial \eta_{xxx}}{\partial t} \right)dx$$
or, more simply,
$$\int\left(\eta_t\eta_{xxx}+ \eta\eta_{xxxt}\right)dx$$

6. Apr 26, 2007

### hanson

Thank you all. I know what to do.
However, here comes new challenge after moving a tiny step forward...
I had a serious trouble in computing the integral in the attachement again. This time is no longer a conceptual problem...