10. 求证: 函数 $f(x)=\frac{x}{x^{2}-1}$ 在 $(1,+\infty)$ 上是减函数.
：证明 证法 1: 设 $x_{1}, x_{2} \in(1,+\infty)$, 且 $x_{1}<x_{2}$,

则 $f\left(x_{1}\right)-f\left(x_{2}\right)=\frac{x_{1}}{x_{1}^{2}-1}-\frac{x_{2}}{x_{2}^{2}-1}$

$=\frac{x_{1}\left(x_{2}^{2}-1\right)-x_{2}\left(x_{1}^{2}-1\right)}{\left(x_{1}^{2}-1\right)\left(x_{2}^{2}-1\right)}$

$=\frac{\left(x_{2}-x_{1}\right)\left(x_{1} x_{2}+1\right)}{\left(x_{1}^{2}-1\right)\left(x_{2}^{2}-1\right)}$.

$\because 1<x_{1}<x_{2}$,

$\therefore x_{2}-x_{1}>0, x_{1} x_{2}+1>0, x_{1}^{2}-1>0, x_{2}^{2}-1>0$.

$\therefore f\left(x_{1}\right)-f\left(x_{2}\right)>0$, 即 $f\left(x_{1}\right)>f\left(x_{2}\right)$.

$\therefore f(x)=\frac{x}{x^{2}-1}$ 在 $(1,+\infty)$ 上是减函数.

证法 2: $\because f(x)=\frac{x}{x^{2}-1}$,

$\therefore f^{\prime}(x)=\frac{x^{2}-1-2 x^{2}}{\left(x^{2}-1\right)^{2}}=\frac{-\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}}$.

$\because x \in(1,+\infty), \quad \therefore f^{\prime}(x)<0$.

$\therefore f(x)=\frac{x}{x^{2}-1}$ 在 $(1,+\infty)$ 上是减函数.