Answer
inside: $g(x)=5 \ln x+7$
outside: $f(g)=4\sin(g(x))+5x$
derivative: $f^{\prime}(x)=\dfrac{20\cos(5\ln x+7) }{x} +5$
Work Step by Step
Given$$
f(x)=4 \sin (5 \ln x+7)+5 x
$$
Use the chain rule to take the derivative
$$
\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)
$$
Here $g(x)=5 \ln x+7$ and $f(g(x))=4\sin(g(x))+5x,$ then
\begin{align*}
f^{\prime}(x) &=\left(4\sin(g(x))+5x\right)^{\prime} \\
&=\left(4\sin(g(x)) \right)^{\prime}+(5x)'\\
&=(4\cos(g(x)))g^{\prime}(x)+5 \\
&=(4\cos(g(x)))\left(\frac{5}{x}\right)+5 \\
&=\frac{20\cos(5\ln x+7) }{x} +5
\end{align*}