Our differential law relies on the change-of-base formula.
\[
\begin{align}
v^u &= \exp (u \mult \log v)
\end{align}
\]
Let's check that this formula holds.
\[
\begin{align}
\exp (u \mult \log v) &= \naturalbase^{u \cdot \log v} = \bigl(\naturalbase^{\log v}\bigr)^u = \bigl(\exp \log v\bigr)^u = v^u
\end{align}
\]
We'll use the change-of-base formula to rewrite the power \( v^u \) and thereby calculate its differential law!
\[
\begin{align}
\diff \bigl(v^u\bigr) &= \diff \bigl(\exp (u \mult \log v)\bigr)
\\
&= \exp (u \mult \log v) \mult \diff (u \mult \log v)
\\
&= v^u \mult \diff (u \mult \log v)
\\
&= v^u \mult \bigl(\log v \mult \diff u + u \mult \diff (\log v)\bigr)
\\
&= v^u \mult \biggl(\log v \mult \diff u + u \mult \frac{1}{v} \mult \diff v\biggr)
\\
&= v^u \mult \log v \mult \diff u + u \mult v^{u - 1} \mult \diff v
\end{align}
\]