解答

$${y=\log\frac{x^2+1}{x^2-1}}$$

$$\begin{eqnarray*} y&=& \log(x^2+1)-\log(x^2-1)\\ y'&=& \frac{(x^2+1)'}{x^2+1}... ...2x\{(x^2-1)-(x^2+1)\}}{(x^2+1)(x^2-1)}\\ &=& -\frac{4x}{x^4-1} \end{eqnarray*}$$