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. $\frac{1}{1+x^2}$
• D. $\frac{-1}{2\left(1+x^2\right)}$

Answer 1

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

Let us take $\frac{f\left(x\right)=\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$

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

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

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

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

$\cos y=\frac{\sqrt{1-\sqrt{1-x^2}}}{\sqrt{2}}$

$-\sin y\frac{dy}{dx}=\frac{1\times \frac{+2x}{2\sqrt{1-x^2}}}{2\sqrt{2}\sqrt{1-\sqrt{1-x^2}}}$

$\frac{-x}{\sqrt{2}\sqrt{1-\sqrt{1-x^2}}}\frac{dy}{dx}=\frac{x}{2\sqrt{2}\sqrt{1-x^2}\sqrt{1-\sqrt{1-x^2}}}$

$\frac{dy}{dx}=\frac{-1}{2}\times \frac{-1}{2}\times \frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=\frac{-1}{2\sqrt{1-x^2}}$

$B$ is correct.