Answer
inside: $g(x)=1+3x$
outside: $f(g)=1+58e^g$
derivative: $f^{\prime}(x)=174 e^{1+3x}$
Work Step by Step
Given$$
f(x)=1+58 e^{(1+3 x)}
$$
Use the chain rule to take the derivative
$$
\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)
$$
Here $g(x)=1+3x$ and $f(g)=1+58e^g,$ then
\begin{align*}
f^{\prime}(x) &=\left(1+58e^g\right)^{\prime} \\
&= 58e^g g^{\prime}(x) \\
&=58(3)e^{1+3x} \\
&=174 e^{1+3x}
\end{align*}