Question

If $y={\sin }^{-1}\frac{2x}{1+x^2}+se{c}^{-1}\frac{1+x^2}{1-x^2}$, then $\frac{dy}{dx}=$

• A. $\frac{4}{\left(1-x^2\right)}$
• B. $\frac{4}{\left(1+x^2\right)}$
• C. $\frac{1}{\left(1+x^2\right)}$
• D. $\frac{-4}{\left(1+x^2\right)}$

Answer 1

The correct option is B

$\frac{4}{\left(1+x^2\right)}$

To find the value of $\frac{dy}{dx}$:

Given,

$\mathrm{y}={\sin }^{-1}\frac{2\mathrm{x}}{1+{\mathrm{x}}^2}+{\sec }^{-1}\frac{1+{\mathrm{x}}^2}{1-{\mathrm{x}}^2}$

This equation can be written as,

$y={\sin }^{-1}\frac{2x}{1+x^2}+{\cos }^{-1}\frac{1-x^2}{1+x^2}\mathbf{}\left[\because co{s}^{-1}x=se{c}^{-1}\frac{1}{x}\right]$

We know, $\frac{d}{dx}{\sin }^{-1}x=\frac{1}{\sqrt{1-x^2}}$ and $\frac{d}{dx}{\cos }^{-1}x=\frac{-1}{\sqrt{1-x^2}}$

Now, differentiate with respect to $x$,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{\sqrt{1-{\left({\displaystyle \frac{2x}{1+x^2}}\right)}^2}}\frac{2\left(1+x^2\right)-\left(2x\right)\left(2x\right)}{{\left(1+x^2\right)}^2}+\frac{-1}{\sqrt{1-{\left({\displaystyle \frac{1-x^2}{1+x^2}}\right)}^2}}\frac{-2x\left(1+x^2\right)-2x\left(1-x^2\right)}{{\left(1+x^2\right)}^2}\\ & =& \frac{\left(1+x^2\right)\left(2-2x^2\right)}{\sqrt{1+x^4+2x^2-4x^2}{\left(1+x^2\right)}^2}+\frac{\left(1+x^2\right)4x}{\sqrt{1+x^4+2x^2-1-x^4+2x^2}{\left(1+x^2\right)}^2}\\ & =& \frac{2\left(1-x^2\right)}{\sqrt{{\left(1-x^2\right)}^2}\left(1+x^2\right)}+\frac{4x}{2x\left(1+x^2\right)}\\ & =& \frac{2}{\left(1+x^2\right)}\left(1+1\right)\\ & =& \frac{4}{\left(1+x^2\right)}\end{array}$

Hence, the correct option is B.