Question

$\frac{d}{dx}\left({\sin }^{-1}\left(\frac{2}{x^{-1}+x}\right)\right)$ is equal to

• A. $\frac{2}{1+x^2},0<|x|<1$
• B. $\frac{2}{\sqrt{1-x^2}},|x|<1$
• C. $\frac{2}{\sqrt{1+x^2}}$
• D. $0$

Answer 1

The correct option is A $\frac{2}{1+x^2},0<|x|<1$

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

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

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

$\Rightarrow \sin y=\frac{2x}{1+x^2}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{(1+x^2)\frac{d}{dx}\left(2x\right)-\left(2x\right)\frac{d}{dx}(1+x^2)}{{(1+x^2)}^2}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{(1+x^2)\times 2-\left(2x\right)\times 2x}{{(1+x^2)}^2}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{2+2x^2-4x^2}{{(1+x^2)}^2}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{2-2x^2}{{(1+x^2)}^2}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{2(1-x^2)}{{(1+x^2)}^2}$

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

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

$\Rightarrow \frac{\sqrt{1+x^4+2x^2-4x^2}}{1+x^2}\frac{dy}{dx}=2\frac{(1-x^2)}{{(1+x^2)}^2}$

$\Rightarrow \frac{\sqrt{1+x^4-2x^2}}{1+x^2}\frac{dy}{dx}=2\frac{(1-x^2)}{{(1+x^2)}^2}$

$\Rightarrow \frac{\sqrt{{(1-x^2)}^2}}{1+x^2}\frac{dy}{dx}=2\frac{(1-x^2)}{{(1+x^2)}^2}$

$\Rightarrow \frac{1-x^2}{1+x^2}\frac{dy}{dx}=2\frac{1-x^2}{{(1+x^2)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{2}{1+x^2}$