# Derivative Simplification

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!

• $\frac {dt}{d \ln(t)}=\frac 1{\frac {d \ln(t)}{dt}}$ could help – Claude Leibovici Dec 10 '18 at 8:35

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