Total Differentials: Taking the Total Differential of Reduced Mass

[eprparadox]

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!

 

[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.