Answer
inside: $g(x)=2^x$
outside: $f(g)= \ln(g(x))$
derivative: $f^{\prime}(x)=\dfrac{2^x\ln 2}{2^x}=\ln 2 $
Work Step by Step
Given$$ f(x)=\ln(2^x) $$
Use the chain rule to take the derivative
$$ \frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x) $$
Here $g(x)= 2^x$ and $f(g)= \ln(g(x)),$ then
\begin{align*}
f^{\prime}(x) &=\left( \ln(g(x))\right)^{\prime} \\
&=\frac{1}{g(x)}g^{\prime}(x) \\
&=\frac{2^x\ln 2}{2^x}\\
&=\ln 2 \end{align*}
