Question

$\frac{d}{dx}\left\{{\cot }^{-1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}=$

• A. $\frac{1}{\sqrt{1-x^2}}$
• B. $\frac{-1}{2\sqrt{1-x^2}}$
• C. $\frac{1}{\sqrt{1+x^2}}$
• D. $\frac{-1}{2\sqrt{1+x^2}}$

Answer 1

The correct option is B $\frac{-1}{2\sqrt{1-x^2}}$

Given,

$\frac{d}{dx}\left({\cot }^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\right)$

$f={\cot }^{-1}\left(u\right),u=\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$

$=\frac{d}{du}\left({\cot }^{-1}\left(u\right)\right)\frac{d}{dx}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

$=(-\frac{1}{u^2+1})\frac{2}{{(\sqrt{1+x}+\sqrt{1-x})}^2\sqrt{x+1}\sqrt{-x+1}}$

$=(-\frac{1}{{\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)}^2+1})\frac{2}{{(\sqrt{1+x}+\sqrt{1-x})}^2\sqrt{x+1}\sqrt{-x+1}}$

$=-\frac{1}{2\sqrt{x+1}\sqrt{-x+1}}$