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Total Differentials

  1. Jul 30, 2014 #1
    Hello!

    I'm reading Mary Boa's "mathematical methods in the physical sciences" and I'm on a section about total differentials.

    So a total differential is for f(x, y) we know to be:

    [tex] df = \frac{\partial f}{\partial x}{dx} + \frac{\partial f}{\partial y}{dy} [/tex]


    Now, I've attached a problem I'm confused about. It involves taking the total differential of the reduced mass equation:

    [tex] \mu^{-1} = m_1^{-1} + m_2^{-2} [/tex]

    In her example, she says to take the total differential of the equation and sets the left side equal to zero. I understand why it's zero (because we want the reduced mass to be unchanged so we want [itex] \partial \mu = 0 [/itex]).

    But essentially, I don't know what it means to just take the differential of [itex] \mu^{-1} [/itex] because I'm accustomed to having some defined function f(x, y) or something and if I take it's differential, I just get [itex] df [/itex].

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Jul 30, 2014 #2

    WWGD

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    I assume the exponent means a^{-1}=1/a. Then d(1/a)=d(a^{-1})=-a^{-1-1}da. This is just d(a^n)=na^{n-1}da.
     
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