Deriving $\frac{8}{\sqrt{x-2}}$ I'm not sure how to derive this:
$$\frac{8}{\sqrt{x-2}}$$

I tried
$$8 \cdot \frac{1}{\sqrt{x-2}}$$
$$8 \cdot (\sqrt{x-2})^{-1}$$
Differentiating w.r.t. $x$,
$$8 \cdot -1 \cdot (\sqrt{x-2})^{-2}$$
$$8 \cdot -1 \cdot \frac{1}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{(\sqrt{x-2})^{2}}$$
$$\frac{-8}{x-2}$$
But the answer is
$$\frac{-4}{(x-2)\sqrt{x-2}}$$

What should I have done?

Since $(\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$, then $$\frac{8}{\sqrt{x-2}}=\frac{0-8(\sqrt{x-2})'}{x-2}=\frac{-4\frac{1}{\sqrt{x-2}}}{x-2}=-\frac{4}{(x-2)\sqrt{x-2}}.$$
