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Need help proving the following:

$\frac{dp}{d ln(t)} = t \frac {dp}{dt} $

This is what I have so far:

Applying chain rule.

$\frac{dp}{d ln(t)} = \frac {dp}{dt}\frac {dt}{d ln(t)} $

Simplifying:

$\frac {dp}{dt}\frac {dt}{d ln(t)} = \frac {dp}{dt}\frac {1}{\frac{d}{dt}} ln(t)$

Need assistance with the rest. Thanks!

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  • $\begingroup$ $\frac {dt}{d \ln(t)}=\frac 1{\frac {d \ln(t)}{dt}}$ could help $\endgroup$ – Claude Leibovici Dec 10 '18 at 8:35
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Leibniz's notations can sometimes be confusing. Writing something like $\frac{1}{\frac{\mathrm{d}}{\mathrm{d}t}}$ is terrible.

Let $p(t) = q(\ln(t))$. Then

$$\frac{\mathrm{d}p(t)}{\mathrm{d}\ln(t)} = \frac{\mathrm{d} q(\ln(t))}{\mathrm{d} \ln(t)} = q'(\ln(t)).$$ Here we treat $\ln(t)$ as a single variable.

Also,

$$\frac{\mathrm{d}p(t)}{\mathrm{d}t} = \frac{\mathrm{d}q(\ln(t))}{\mathrm{d}t}= q'(\ln(t))(\ln(t))'=q'(\ln(t))\frac{1}{t}.$$

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