# Total Differentials

1. Jul 30, 2014

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:

$$df = \frac{\partial f}{\partial x}{dx} + \frac{\partial f}{\partial y}{dy}$$

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

$$\mu^{-1} = m_1^{-1} + m_2^{-2}$$

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 $\partial \mu = 0$).

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

Thanks!

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2. Jul 30, 2014

### WWGD

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.