合成関数の微分公式から微分する問題

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

$${\hspace{10pt}\left\{\frac{f(x)}{g(x)}\right\}' = \frac{f'(x)g(x)-f(x)g'(x)}{\{g(x)\}^2}\hspace{10pt}}$$

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