Calculus 10th Edition

$\displaystyle \frac{1}{1-x^{2}}$
By theorem 5.3/2:$\ \ \ \displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$ --------------- Here,$u(x)=\displaystyle \sqrt{\frac{x+1}{x-1}}=\frac{(x+1)^{1/2}}{(x-1)^{1/2}}$ For $u^{\prime}(x)$ we use the quotient rule: $u^{\prime}(x)=\displaystyle \frac{[\frac{1}{2}(x+1)^{-1/2}\cdot 1]\cdot(x-1)^{1/2}-(x+1)^{1/2}\cdot[\frac{1}{2}(x-1)^{-1/2}\cdot 1]}{[(x-1)^{1/2}]^{2}}$ $=\displaystyle \frac{\sqrt{\frac{x-1}{x+1}}-\sqrt{\frac{x+1}{x-1}}}{2(x-1)}\cdot\frac{\sqrt{(x+1)(x-1)}}{\sqrt{(x+1)(x-1)}}=\frac{(x-1)-(x+1)}{2(x-1)\sqrt{(x+1)(x-1)}}$ $=\displaystyle \frac{-2}{2(x-1)\sqrt{(x+1)(x-1)}}=-\frac{1}{(x-1)\sqrt{(x+1)(x-1)}}$ So $\ \ \ \displaystyle \frac{d}{dx}[\ln\sqrt{\frac{x+1}{x-1}}]=\frac{1}{\sqrt{\frac{x+1}{x-1}}}\cdot\frac{-1}{(x-1)\sqrt{(x+1)(x-1)}}$ $=-\displaystyle \frac{1}{(x-1)(x+1)}=\frac{-1}{x^{2}-1}=\frac{1}{1-x^{2}}$