# Differentiation trick

1. May 12, 2005

### Baggio

I keep seeing this trick everywhere but I don't see how it is done.

how do we go from

[itex]\frac{1}{r^{2}}\frac{d}{dr}(r^{2}\frac{d \rho}{dr} [\latex]

to

[itex]\frac{1}{r} \frac{d^{2} \rhor}{dr^2{}}[\latex]

Ugggh can't get latex to work anyway it's

(1/r^2)(d/dr(r^2. dx/dr)

how do we go from that to

(1/r)(d/dr(r^2.d(xr)/dr))

I know for sure that they're equal I just don't know how to manipulate it! :(

Thanks

Last edited: May 12, 2005
2. May 12, 2005

### dextercioby

Use [ tex ] [ /tex ] (without the spaces).

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

and u can see pretty clearly the 2 things are different.

Daniel.

3. May 12, 2005

### dextercioby

Who's "x" and what does he do...?

Daniel.

4. May 12, 2005

### Baggio

No, let's start again

(1/r^2)(d/dr(r^2. dx/dr)

&

(1/r)(d/dr(r^2.d(xr)/dr))

x is a function of r, if you expand both you get the same result, but how can I go directly from the top eq to the bottom