# I can't believe the hint that is provided; can someone explain how it is true?

1. Feb 20, 2012

### Tsunoyukami

I'm working on a question for a problem set that has the following hint:

For any function A:

$\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})$

I don't understand how this is true; this is how I try to show that both sides are equal but it doesn't work out...any help would be appreciated.

$\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})$

$\frac{d}{dr}\frac{dA}{dr} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})$

$(\frac{d}{dr} + \frac{2}{r}) \frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})$

$\frac{d}{dr} + \frac{2}{r} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2})$

$\frac{d}{dr} + \frac{2}{r} = \frac{d}{dr}$

$\frac{2}{r} = \frac{d}{dr} - \frac{d}{dr}$

$\frac{2}{r} = 0$

This is only true if r $\rightarrow$ $\infty$. I think I must be doing something wrong - can anyone please explain what I'm doing wrong and show me how the hint is a true statement? Thanks!

2. Feb 20, 2012

### Mindscrape

You're treating everything as if it's multiplication, that's not legal.

Last edited by a moderator: Feb 20, 2012
3. Feb 20, 2012

### Staff: Mentor

The step above makes no sense. d/dr(r2) does NOT mean d/dr * r2, so you can't divide by r2 to end up with d/dr.

There are steps above that that also don't make sense, for essentially the same reason.
The assumption here is that A = A(r); i.e., A is a function of r. Also, it is assumed that A is twice differentiable.

Start on the right side and calculate (1/r2) * d/dr(r2 *dA/dr).

4. Feb 20, 2012

### Tsunoyukami

Oh man, I can't believe I missed that - thanks so much it makes perfect sense now! It slipped my mind that the $\frac{d}{dr}$ would act on the $r^{2}$ and that I would have to use the chain rule. Thanks a lot!

I see my mistake, thank you. I knew I was doing something fishy but I wasn't certain what it was. Thank you again for the prompt replies!