If y = sin^{-1}left( frac{1 - x^2}{1 + x^2} r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} ight)$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting this into the expressio

Vedclass · Accepted Answer

$\pm \frac{2}{1 + x^2}$

Vedclass · Answer

(C) Given $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} ight)$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting this into the expression,we get $y = \sin^{-1}(\cos 2	heta)$.Since $\cos 2	heta = \sin\left( \frac{\pi}{2} - 2	heta ight)$ or $\sin\left( 2	heta - \frac{\pi}{2} ight)$,we have $y = \frac{\pi}{2} \pm 2	heta$.Substituting $	heta = 	an^{-1} x$,we get $y = \frac{\pi}{2} \pm 2 	an^{-1} x$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \pm 2 \cdot \frac{1}{1 + x^2} = \pm \frac{2}{1 + x^2}$.

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

Similar Questions

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

The derivative of ${\tan ^{ - 1}}\left( {\frac{{\sin x - \cos x}}{{\sin x + \cos x}}} \right)$ with respect to $\frac{x}{2}$,where $x \in \left( {0, \frac{\pi }{2}} \right)$,is

If $y = \sin^{-1}\left(\frac{3x}{2} - \frac{x^3}{2}\right)$,then $\frac{dy}{dx}$ is equal to

If $a > b > 0$ and $x$ is acute,then $\frac{d}{dx} \left[ \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} \right) \right] = $

Find the derivative of $\cos^{-1} \sqrt{\frac{1 + x^2}{2}}$ with respect to $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module