Why does $\mathrm{d}l=v\,\mathrm{d}t$ imply that $\frac{\mathrm{d}}{\mathrm{d}t}=v\frac{\mathrm{d}}{\mathrm{d}l}$? If we have a lengthening pendulum and the length $l$ of the pendulum at time $t$ is $$l=l_0+vt$$ where $l_0$ is the initial length of the pendulum and $v$ is the velocity for which the pendulum's length is increasing.
Then the differential change in the length of the pendulum is given by $$\mathrm{d}l=v\,\mathrm{d}t$$ this implies that $$\frac{1}{\mathrm{d}t}=v\frac{1}{\mathrm{d}l}\tag{1}$$ My question is why does it now follow that $$\frac{\mathrm{d}}{\mathrm{d}t}=v\frac{\mathrm{d}}{\mathrm{d}l}$$ Do we simply multiply both sides of $(1)$ by $\mathrm{d}$ or is there more to it than that? 
 A: The chain rule gives
$$\begin{align}
\left(\frac{d}{dt}\right)f(l(t))&=\frac{df(l)}{dl}\frac{dl}{dt}\\\\
&=\left(v\frac{d}{dl}\right)f(l)
\end{align}$$
A: Here is an expansion on the above answer. Technically speaking, it doesn't make sense to "multiply" both sides by $d$ or "divide" by infintesimals (without proper justification). What you should really say is that a small change $\Delta l$ is related to a small change in $t$ via
$$\Delta l=v\Delta t\;.$$
The quantities $\Delta l$ and $\Delta t$ are positive quantities which can be made arbitrarily small, but are nonzero. More precisely, we have not yet taken the limit $\Delta t\to 0$. To get the equation you are looking for, we should see how taking the derivative of a function $f$ of $l$ relates to taking the derivative with respect to $t$. Let us assume that $l$ is a function of $t$ and let us pick a point $t_0$. By definition, we have
$$\frac{df}{dt}(l(t_0))=\lim_{\Delta t\to 0}\frac{f(l(t))-f(l(t_0))}{\Delta t}=\lim_{\Delta t\to 0}\frac{f(l(t))-f(l(t_0))}{\Delta l}\cdot \frac{\Delta l}{\Delta t}$$$$=\lim_{\Delta t\to 0}\frac{f(l(t))-f(l(t_0))}{\Delta l}\cdot v$$
Since $\Delta l\to 0$ as $\Delta t \to 0$, the RHS gives you exactly 
$v\dfrac{df}{dl}(l_0)$. Since the point $t_0$ was arbitrary, we get the desired equation
$$\frac{df}{dt}=v\frac{df}{dl}\;.$$
Since the function $f$ was arbitrary, the differential operators are related via the equation
$$\frac{d}{dt}=v\frac{d}{dl}\;.$$
I have sort of implicitly used the proof of the chain rule, but I wanted to provide a solution which more closely follows your thought process.
