Answer
inside: $g(x)=5 x^{2}+3 x+7$
outside: $f(g)=g^{-1}$
derivative: $f^{\prime}(x)=-\left(5 x^{2}+3 x+7\right)^{-2}(10x+3 )$
Work Step by Step
Given $$
f(x)=\left(5 x^{2}+3 x+7\right)^{-1}
$$
Use the chain rule to take the derivative
$$\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)$$
Here $g(x)= 5 x^{2}+3 x+7$ and $ f(g )=g^{-1}$, then
\begin{align*}
f'(x)&=(g^{-1}(x))'\\
&=-g^{-2}(x)g'(x)\\
&=-\left(5 x^{2}+3 x+7\right)^{-2}(10x+3 )
\end{align*}
