10
$\begingroup$

I'm trying to figure out how to take a derivative that looks like $\displaystyle \frac{d}{d(\ln(a))}$, of a function $F(a)$, where $a = a(t)$. In the paper I'm reading (where this appears), they give the following result in the case that $F(a) = \frac{\dot{a}}{a}$ (where the "dot" is a derivative with respect to $t$):

$\frac{d(1/F^2)}{d\ln(a)} = \frac{-2\dot{F}}{F^4},$

but I can't see how they're getting this. Any insight would be much appreciated.

  • 0
    Possible duplicate: https://math.stackexchange.com/questions/291376/differentiate-with-respect-to-a-function2017-12-07

1 Answers 1

12

That is simply the chain rule. Let $y(t)=\ln(a(t))$, so that $dy/dt = \dot a/a=F$. Then $\frac{d(F^{-2})}{dt}=\frac{d(F^{-2})}{dy}\frac{dy}{dt},$

hence

$-2\dot F F^{-3} = \frac{d(F^{-2})}{dy} \frac{\dot a}{a} = \frac{d(F^{-2})}{dy} F,$

from which $\frac{d(F^{-2})}{dy} = -2\dot F F^{-4}.$