Using the chain rule:
d d x { x 2 + 1 x 2 − 1 } = 1 2 x 2 + 1 x 2 − 1 ⋅ d d x { x 2 + 1 x 2 − 1 } = 1 2 x 2 − 1 x 2 + 1 ⋅ d d x { x 2 + 1 x 2 − 1 } And now, the quotient rule: = 1 2 x 2 − 1 x 2 + 1 ⋅ ( ( x 2 − 1 ) ( 2 x ) − ( x 2 + 1 ) 2 x ( x 2 − 1 ) 2 ) = 1 2 x 2 − 1 x 2 + 1 ⋅ ( − 4 x ( x 2 − 1 ) 2 ) = x 2 − 1 x 2 + 1 ⋅ ( − 2 x ( x 2 − 1 ) 2 ) = − 2 x ( x 2 − 1 ) x 4 − 1 \begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sqrt{\frac{x^2+1}{x^2-1}}\right\}&=\frac{1}{2\sqrt{ \frac{x^2+1}{x^2-1}}}\cdot \frac{\mathrm{d}}{\mathrm{d}x}\left\{ \frac{x^2+1}{x^2-1}\right\}\\
&=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{x^2+1}{x^2-1}\right\}
\\ \text{And now, the quotient rule:} \\
&=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{(x^2-1)(2x)-(x^2+1)2x}{(x^2-1)^2}\right)\\
&=\frac{1}{2}\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{-4x}{(x^2-1)^2}\right)\\
&=\sqrt{\frac{x^2-1}{x^2+1}}\cdot\left(\frac{-2x}{(x^2-1)^2}\right)\\
&=\frac{-2x}{({x^2-1})\sqrt{x^4-1}}
\end{aligned} d x d { x 2 − 1 x 2 + 1 } And now, the quotient rule: = 2 x 2 − 1 x 2 + 1 1 ⋅ d x d { x 2 − 1 x 2 + 1 } = 2 1 x 2 + 1 x 2 − 1 ⋅ d x d { x 2 − 1 x 2 + 1 } = 2 1 x 2 + 1 x 2 − 1 ⋅ ( ( x 2 − 1 ) 2 ( x 2 − 1 ) ( 2 x ) − ( x 2 + 1 ) 2 x ) = 2 1 x 2 + 1 x 2 − 1 ⋅ ( ( x 2 − 1 ) 2 − 4 x ) = x 2 + 1 x 2 − 1 ⋅ ( ( x 2 − 1 ) 2 − 2 x ) = ( x 2 − 1 ) x 4 − 1 − 2 x