Chain rule with a base-2 exponential composite function

Write $f(x) = 2^{u(x)}$ where the inner function is $u(x) = x^2 - 1$. For a base-$a$ exponential, the general differentiation rule is $\frac{d}{dx}[a^{u}] = a^{u} \ln(a) \cdot u'(x)$, which comes from writing $a^u = e^{u\ln a}$ and applying the chain rule with the exponential rule $\frac{d}{dx}[e^x] = e^x$.

Here $a = 2$ and $u'(x) = \frac{d}{dx}[x^2 - 1] = 2x$. Substituting into the rule gives $f'(x) = 2^{x^2-1} \ln(2) \cdot (2x)$, which simplifies to $f'(x) = 2x \ln(2) \cdot 2^{x^2 - 1}$.