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? 

The chain rule gives
$$\begin{align}
\left(\frac{d}{dt}\right)f(l(t))&=\frac{df(l)}{dl}\frac{dl}{dt}\\\\
&=\left(v\frac{d}{dl}\right)f(l)
\end{align}$$
