let $f$ be a differentiable function. Compute $\frac{d}{dx}g(2)$, where $g(x) = \frac{f(2x)}{x}$. let $f$ be a differentiable function and  $$\lim_{x\to 4}\dfrac{f(x)+7}{x-4}=\dfrac{-3}{2}.$$
Define $g(x)=\dfrac{f(2x)}{x}$. I want to know the derivative
$$\dfrac{d}{dx}g(2)=?$$

I know that :
$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$
and :
$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}=a\in\mathbb{R}$$
so :
$$\lim_{x\to 4}\dfrac{f(x)-f(4)+7+f(4)}{x-4}=\dfrac{-3}{2}$$
$$\lim_{x\to 4}\dfrac{f(x)-f(4)}{x-4}+\dfrac{f(4)+7}{x-4}=\dfrac{-3}{2}$$
now what ?

HINT
\begin{align*}
\frac{dg}{dx}(2) & = \lim_{x\rightarrow 2} = \frac{g(x) - g(2)}{x - 2} = \lim_{x\rightarrow 2}\frac{\displaystyle\frac{f(2x)}{x} - \frac{f(4)}{2}}{x - 2} = \lim_{x\rightarrow 2}\frac{2f(2x)-xf(4)}{2x(x-2)}\\
& = \lim_{u\rightarrow 4}\frac{2f(u) - u\displaystyle\frac{f(4)}{2}}{u\left(\displaystyle\frac{u}{2}-2\right)} = \lim_{u\rightarrow 4}\frac{4f(u)-uf(4)}{u(u-4)} = \lim_{u\rightarrow 4}\frac{4f(u)-uf(4)}{u^2 - 4u}\\
& \overset{\mathrm{L'H}}{=}\lim_{u\rightarrow 4}\frac{4f'(u)-f(4)}{2u-4} = \frac{4f'(4)-f(4)}{4}
\end{align*}
