Question

If $y={\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{dy}{dx}=$

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

Answer 1

The correct option is C

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

Finding the value of $\frac{dy}{dx}$:

Given,

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

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

So,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{\sqrt{1-{\left(\frac{1-x^2}{1+x^2}\right)}^2}}\frac{\left(-2x\left(1+x^2\right)-2x\left(1-x^2\right)\right)}{{\left(1+x^2\right)}^2}\\ & =& \frac{1+x^2}{\sqrt{1+x^4+2x^2-1-x^4+2x^2}}\frac{\left(-2x-2x^3-2x+2x^3\right)}{{\left(1+x^2\right)}^2}\\ & =& \frac{-4x}{2x\left(1+x^2\right)}\\ & =& \frac{-2}{\left(1+x^2\right)}\end{array}$

Hence, the correct option is C.