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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.

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I'm not fond of this notation. If pressed, I would instead define some new variable, say, $v=\ln\,a$ and then express $1/F^2$ in terms of $v$... –  Guess who it is. Aug 1 '11 at 5:53

1 Answer 1

up vote 10 down vote accepted

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


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

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

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