<s>[INST] 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 ?
 [/INST] You have an extra $4$ in the numerator here:

i know that :
$$\dfrac{d}{dx}g(2)=\dfrac{4(\dfrac{d}{dx}f(4))-4f(4)}{4}$$

If $g(x) = \dfrac{f(2x)}x$, then
\begin{align*}
  \frac d{dx} g(x) &= \frac d{dx} \left(\frac{f(2x)}x\right)\\[0.3cm]
  &= \frac{x \frac d{dx} f(2x) - f(2x) \frac d{dx}x}{x^2}\\[0.3cm]
  &= \frac{2x f'(2x) - f(2x)}{x^2}\\[0.3cm]
\end{align*}
By $\dfrac d{dx} g(2)$ I think you mean $\dfrac d{dx} g(x)\bigg|_{x=2}$.  Or, more compactly, $g'(2)$.
So $g'(2) = \dfrac{4f'(4) - f(4)}4$.
We're given $\displaystyle \lim_{x\to 4} \frac{f(x) + 7}{x-4} = -\frac32$.  Since $\lim_{x\to4} (x-4) = 4-4 = 0$, then $\displaystyle\lim_{x\to4} (f(x) + 7)$ must also be zero because this is the only way the limit of the quotient can be a finite number.  Also, since $f$ is continuous (because it is differentiable) then we can safely say $\displaystyle\lim_{x\to 4}(f(x) + 7) = f(4) + 7$.  Thus, $f(4) + 7 = 0$, and so we have $f(4) = -7$.
It remains to find $f'(4)$.  By the definition of the derivative (sometimes called the alternate definition of the derivative), we have:
$$ f'(4) = \lim_{x \to 4} \frac{f(x) - f(4)}{x-4} = \lim_{x \to 4} \frac{f(x) + 7}{x - 4}$$
And we were given that this is $-\dfrac32$.
Now plug all the numbers into where they need to go, and simplify.
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