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Differentiation trick

  1. May 12, 2005 #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]

    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. jcsd
  3. May 12, 2005 #2

    dextercioby

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    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)
    =\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.
     
  4. May 12, 2005 #3

    dextercioby

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

    Daniel.
     
  5. May 12, 2005 #4
    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
     
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