<s>[INST] 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?
 [/INST] $$\frac{d(f(g(x)))}{dx}\ne f'(g(x))\quad\text{instead}\quad \frac{d(f(g(x)))}{dx}= f'(g(x))g'(x)$$
So
$$(f(\sqrt{x-2}))'=f'(\sqrt{x-2})(\sqrt{x-2})'=f'(\sqrt{x-2})\frac1{2(\sqrt{x-2})}$$
When $f(x)=8/x$ then $f'(x)=-8/x^2$
$$f'(\sqrt{x-2})=\frac{-8}{(\sqrt{x-2})^2}\frac1{2(\sqrt{x-2})}$$
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