3√(x^2-2x)を微分する問題

$${\hspace{10pt}\{f(\hspace{1pt}g(x))\}' = f'(\hspace{1pt}g(x))\hspace{1pt}g'(x)\hspace{10pt}}$$

$$\begin{aligned} \hspace{10pt}y' & =\frac{1}{3} (x^2-2x)^{-\frac{2}{3}} \{x^2-2x\}'\hspace{10pt}\\[1em] & = \frac{1}{3} (x^2-2x)^{-\frac{2}{3}} (2x-2)\\[1em] & = \frac{2(x-1)}{3\sqrt[3]{(x^2-2x)^2}} \\[1em] \end{aligned}$$