# Landau and Lifschitz derivation question

1. Oct 27, 2013

### HJ Farnsworth

Greetings,

On page 7 of Landau and Lifschitz Vol. 1 3rd. Ed, it says

$L(v'^2)=L(v^2+2\bf{v}\cdot\bf{ε}+\bf{ε}^2)$.

They then Taylor expand in powers of ε, getting (ignoring second order terms and higher)

$L(v'^2)=L(v^2)+\frac{\partial L}{\partial v^2}2\bf{v}\cdot\bf{ε}$.

The $\frac{\partial L}{\partial v^2}$ confuses me. We effectively have a function of the form

$f(g(x))=f(a+bx+cx^2)$, which, Taylor expanding around $x=0$, would give

$f(a+bx+cx^2)=f(x=0)+\frac{df}{dg}(x=0)\frac{dg}{dx}(x=0)x=\frac{df}{dg}(x=0)bx$.

So, where I have $\frac{df}{dg}(x=0)$, Landau has $\frac{\partial f}{\partial a}$.

How do I go from what I have to what Laundau has?

Thanks.

-HJ Farnsworth

2. Oct 27, 2013

### dauto

Landau's notation is just a short hand for your notation so there isn't any math to explain here.