What is the result of $\displaystyle \frac{d}{dx}\left(x\frac{d}{dx}\right)$?

Do you use the product rule or is it simply $\displaystyle x\frac{d^2}{dx^2}$?

Thanks

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- Apr 8th 2013, 01:32 PMGeo877Small question about differential operators
What is the result of $\displaystyle \frac{d}{dx}\left(x\frac{d}{dx}\right)$?

Do you use the product rule or is it simply $\displaystyle x\frac{d^2}{dx^2}$?

Thanks - Apr 8th 2013, 01:43 PMPlatoRe: Small question about differential operators
- Apr 9th 2013, 07:08 AMGeo877Re: Small question about differential operators
Thanks Plato, I can't say I'm a big fan of it, but it's rather ubiquitous in physics. So the correct result is:

$\displaystyle x\frac{d^2}{dx^2}+\frac{d}{dx}$ - Apr 9th 2013, 07:56 AMtopsquarkRe: Small question about differential operators
If you are looking at this as a Physicist the typical way to introduce this kind of concept is to use your operator on a function. For example:

$\displaystyle \frac{d}{dx} \left ( x \frac{d}{dx} \right ) f = \frac{d}{dx} \left ( x \frac{df}{dx} \right )$

Use the chain rule:

$\displaystyle \frac{d}{dx} \left ( x \frac{d}{dx} \right ) f = \frac{df}{dx} + x \frac{d^2f}{dx^2}$

And now "factor" the f out:

$\displaystyle \frac{d}{dx} \left ( x \frac{d}{dx} \right ) f = \left ( \frac{d}{dx} + x \frac{d^2}{dx^2} \right ) f$

Thus

$\displaystyle \frac{d}{dx} \left ( x \frac{d}{dx} \right ) = \frac{d}{dx} + x \frac{d^2}{dx^2}$

This is often used for QM and perhaps Mathematical Physics classes until the students learn to "grok" doing this without the function. I admit that the method I did above is little different from just applying the chain rule but Physicists often use it to keep their heads on straight. (I learned it myself, not from one of my Physics classes, but in a text I inherited from a former Professor on Laplace transforms.)

-Dan