Answer
inside: $g(x)=2x+5$
outside: $f(g)=3\sin(g(x))+7$
derivative: $f^{\prime}(x)=6\cos (2x+5)$
Work Step by Step
Given$$
f(x)=3 \sin (2 x+5)+7
$$
Use the chain rule to take the derivative
$$
\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)
$$
Here $g(x)=2x+5$ and $f(g)=3\sin(g(x))+7,$ then
\begin{align*}
f^{\prime}(x) &=\left(3\sin(g(x))+7\right)^{\prime} \\
&=(3\cos(g(x)))g^{\prime}(x) \\
&=(3\cos(g(x)))(2) \\
&=6\cos (2x+5)
\end{align*}
