Answer
inside: $g(x)=4 x+7$
outside: $f(g)=350g^{-1}$
derivative: $f^{\prime}(x)=-1400\left(4 x+7\right)^{-2}=\dfrac{-1400}{(4x+7)^2}$
Work Step by Step
Given $$
f(x)=\frac{350}{4 x+7}
$$
Rewriting $f(x) $ as$$
f(x)= 350(4 x+7)^{-1}
$$
Use the chain rule to take the derivative
$$\frac{d f(g(x))}{d x}=f^{\prime}(g(x)) g^{\prime}(x)$$
Here $g(x)=4 x+7$ and $ f(g )=350g^{-1}$, then
\begin{align*}
f'(x)&=(350g^{-1}(x))'\\
&=-350g^{-2}(x)g'(x)\\
&=-350\left(4 x+7\right)^{-2}(4 )\\
&=-1400\left(4 x+7\right)^{-2}\\
&=\frac{-1400}{(4x+7)^2}
\end{align*}
