Answer
inside: $g(x)=3\ln x-7$
outside: $f(g)= 5 \sin (g(x))-4$
derivative: $f^{\prime}(x)=\dfrac{15\cos ( 3\ln x-7)}{x}$
Work Step by Step
Given
$$ f(x)=5 \sin (3 \ln x-7)-4 $$
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)= 3 \ln x-7 $ and $f(g)=5 \sin (g(x))-4,$ then
\begin{align*}
f^{\prime}(x) &=\left( 5 \sin (g(x))-4\right)^{\prime} \\
&=(5\cos (g(x)))g^{\prime}(x) \\
&=(5\cos (g(x))) \frac{3}{x}\\
&= \frac{15\cos ( 3\ln x-7)}{x}
\end{align*}
