How to show the chain rule using these Leibniz's notations If $x=e^t$, show that (using the chain rule)
$$\frac{d}{dx} = e^{-t}\frac{d}{dt}$$
and
$$\frac{d^2}{dx^2} = e^{-2t}\left[\frac{d^2}{dt^2}-\frac{d}{dt}\right].$$
 A: If $x=e^{t}$, then by the chain rule we have
$$\frac{d}{dx}\{\cdot\}=\frac{1}{\frac{dx}{dt}}\frac{d}{dt}\{\cdot\} =e^{-t}\frac{d}{dt}\{\cdot\}$$
For the second derivative, we have
$$\begin{align}
\frac{d^2}{dx^2}\{\cdot\}&=\frac{d}{dx}\left(\frac{d}{dx}\{\cdot\}\right)\\\\
&=\frac{d}{dx}\left(e^{-t}\frac{d}{dt}\{\cdot\}\right)\\\\
&e^{-t}\frac{d}{dt}\left(e^{-t}\frac{d}{dt}\{\cdot\}\right)\\\\
&=e^{-2t}\frac{d^2}{dt^2}\{\cdot\}-e^{-2t}\frac{d}{dt}\{\cdot\}
\end{align}$$
A: The substitution $x=e^t$ is typically used when solving Cauchy-Euler equations.
If $x=e^t$ then $dx=e^tdt$ so $\dfrac{dt}{dx}=e^{-t}$.
Using the chain rule, 
$$\dfrac{dy}{dx}=\dfrac{dt}{dx}\cdot\dfrac{dy}{dt}=e^{-t}\frac{dy}{dt}\tag{1}$$
Then for the second derivative we have
\begin{eqnarray}
\frac{d^2y}{dx^2}&=&\frac{d}{dx}\left(\frac{dy}{dx}\right)\\
&=&e^{-t}\left(\frac{d}{dt}\left(e^{-t}\dfrac{dy}{dt}\right)\right) \text{ using equation } (1)\\
&=&e^{-t}\left(e^{-t}\frac{d^2y}{dt^2}-e^{-t}\frac{dy}{dt}\right)\\
&=&e^{-2t}\left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)
\end{eqnarray}
