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

[Tsunoyukami]

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

For any function A:

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

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.

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr}\frac{dA}{dr} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex](\frac{d}{dr} + \frac{2}{r}) \frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{d}{dr}[/itex]

[itex] \frac{2}{r} = \frac{d}{dr} - \frac{d}{dr}[/itex]

[itex] \frac{2}{r} = 0[/itex]

This is only true if r [itex]\rightarrow[/itex] [itex]\infty[/itex]. 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!

 

[Mindscrape]

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

 

[Mark44]

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

For any function A:

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

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.

[itex]\frac{d^{2}A}{dr^{2}} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr}\frac{dA}{dr} + \frac{2}{r}\frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex](\frac{d}{dr} + \frac{2}{r}) \frac{dA}{dr} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2} \frac{dA}{dr})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{1}{r^{2}}\frac{d}{dr}(r^{2})[/itex]

[itex]\frac{d}{dr} + \frac{2}{r} = \frac{d}{dr}[/itex]

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.

[itex] \frac{2}{r} = \frac{d}{dr} - \frac{d}{dr}[/itex]

[itex] \frac{2}{r} = 0[/itex]

This is only true if r [itex]\rightarrow[/itex] [itex]\infty[/itex]. 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!

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).

 

[Tsunoyukami]

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

[tex]\frac{1}{r^2}\frac{d}{dr}(r^2 \frac{dA}{dr}) =\frac{1}{r^2}( 2r\frac{dA}{dr}+r^2\frac{d^2A}{dr^2})[/tex]

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

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).

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!