$\frac{d}{dx}\{{\mathrm{Cot}}^{-1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\}=$

$\text{Hint: Put}x=\cos \theta$

${\mathrm{Cot}}^{-1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}={\mathrm{Cot}}^{-1}\frac{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}$

${\mathrm{Cot}}^{-1}\frac{\sqrt{2}\cos {\displaystyle \frac{\theta }{2}}-\sqrt{2}\sin \frac{\theta }{2}}{\sqrt{2}\cos \frac{\theta }{2}+\sqrt{2}\sin \frac{\theta }{2}}={\mathrm{Cot}}^{-1}\frac{\cos {\displaystyle \frac{\theta }{2}}-\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}=co{t}^{-1}cot\left(\frac{\mathrm{\pi }}{4}+\frac{\theta }{2}\right)$

$\frac{d}{dx}\{{\mathrm{Cot}}^{-1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\}=\frac{d}{dx}\{\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1}x\}=\frac{-1}{2\sqrt{1-x^2}}$