Question on the Derivation of Temperature/Scale Factor Relation (2025)

In 'Introduction to Cosmology' by Barbara Ryden, there is an argument made using the first law of thermodynamics to derive the relation T(t) ∝ a(t)^-1 on pages 29 and 30.
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I've been able to work out all the omitted details up to 2.37, which gives the following relation:

$$ \frac{1}{T} \frac{dT}{dt} = - \frac{1}{3V} \frac{dV}{dt}$$

She informs the reader that since V ∝ a(t)³, the rate of change of the photons' temperature is related to the rate of expansion of the universe by the following relation (2.38):

$$ \frac{d}{dt} (ln\,T) = - \frac{d}{dt} (ln\,a) $$

I'm running into a little difficulty seeing how she arrives at 2.38. Here's my attempt getting there from 2.37:

$$ \frac{d}{dt} (\frac{dT}{T}) = -\frac{1}{3V} (\frac{d}{dt}\, a(t)^3)$$

$$ \frac{d}{dt} (\frac{dT}{T}) = -\frac{1}{3a(t)^3} ( 3a(t)^2)$$

$$ \frac{d}{dt} (\frac{dT}{T}) = -\frac{1}{a(t)}$$

$$ \frac{d}{dt} (\int \frac{dT}{T}) = -\frac{1}{a(t)}$$

$$ \frac{d}{dt} (ln\,T) = -\frac{1}{a(t)}$$

I can't seem to get the d/dt piece on the right-hand side like at the top of my post. I thought maybe I could pull out the d/dt from the expression, like the following:

$$- \frac{1}{3V} \frac{dV}{dt} = -\frac{1}{3} \frac{d}{dt}(\frac{dV}{V})$$

But that doesn't seem to net me the a(t)^-1 result like my original path did. Is there some simple mistake I'm making in my calculations here? Additionally, is the negative sign "absorbed" into the proportionality expression? It doesn't seem to make much sense that I'm left with a negative sign in what I'm doing, since the scale factor never has a negative value.

Thank you very much for your time and any assistance you may be able to provide.

You left out the chain rule on the second step.

TRB8985 said:

$$ \frac{d}{dt} (\frac{dT}{T}) = -\frac{1}{3V} (\frac{d}{dt}\, a(t)^3)$$

The far right term simplifies to:

$${d \over dt} a(t)^3 = 3 a(t)^2 {da \over dt}$$

Leaving the chain rule expansion in place should make the rest of the derivation fall into place nicely.