Simplifying (1/r²)d/dr(r² dρ/dr)

  • Context: Undergrad 
  • Thread starter Thread starter Baggio
  • Start date Start date
  • Tags Tags
    Differentiation
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 3K views
Baggio
Messages
211
Reaction score
1
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]<br /> <br /> to<br /> <br /> [itex]\frac{1}{r} \frac{d^{2} \rhor}{dr^2{}}[\latex]<br /> <br /> Ugggh can't get latex to work anyway it's<br /> <br /> (1/r^2)(d/dr(r^2. dx/dr)<br /> <br /> how do we go from that to<br /> <br /> (1/r)(d/dr(r^2.d(xr)/dr))<br /> <br /> I know for sure that they're equal I just don't know how to manipulate it! :(<br /> <br /> Thanks[/itex][/itex]
 
Last edited:
Physics news on Phys.org
Use [ tex ] [ /tex ] (without the spaces).

[tex]\frac{1}{r^{2}}\left(2r\frac{d\rho}{dr}+r^{2}\frac{d^{2}\rho}{dr^{2}}\right)<br /> =\frac{2}{r}\frac{d\rho}{dr}+\frac{d^{2}\rho}{dr}[/tex]

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

Daniel.
 
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