The derivative of y = sin^{-1}left(frac{sqrt{ | vedclass

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

Question

(D) $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}ight)$ Let $x = \cos 2	heta$,then $2	heta = \cos^{-1} x$ or $	heta = \frac{1}{2} \cos^{-

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{1-x^2}}$

Vedclass · Answer

(D) $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}ight)$ Let $x = \cos 2	heta$,then $2	heta = \cos^{-1} x$ or $	heta = \frac{1}{2} \cos^{-1} x$. Substituting $x$ in the expression: $y = \sin^{-1}\left(\frac{\sqrt{1+\cos 2	heta} - \sqrt{1-\cos 2	heta}}{2}ight)$ Using trigonometric identities $1+\cos 2	heta = 2\cos^2 	heta$ and $1-\cos 2	heta = 2\sin^2 	heta$: $y = \sin^{-1}\left(\frac{\sqrt{2}\cos 	heta - \sqrt{2}\sin 	heta}{2}ight) = \sin^{-1}\left(\frac{\cos 	heta - \sin 	heta}{\sqrt{2}}ight)$ $y = \sin^{-1}\left(\frac{1}{\sqrt{2}}\cos 	heta - \frac{1}{\sqrt{2}}\sin 	hetaight)$ $y = \sin^{-1}\left(\sin \frac{\pi}{4} \cos 	heta - \cos \frac{\pi}{4} \sin 	hetaight)$ $y = \sin^{-1}\left(\sin\left(\frac{\pi}{4} - 	hetaight)ight) = \frac{\pi}{4} - 	heta$ Substituting $	heta = \frac{1}{2} \cos^{-1} x$: $y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$ Differentiating with respect to $x$: $\frac{dy}{dx} = 0 - \frac{1}{2} \left(-\frac{1}{\sqrt{1-x^2}}ight) = \frac{1}{2\sqrt{1-x^2}}$ Thus,option $D$ is correct.

The derivative of $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)$ is

Similar Questions

If $y = \cot^{-1}(\cos 2x)^{1/2}$,then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is:

Derivative of $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ with respect to $\cos ^{-1}\left(4 x^3-3 x\right)$ is

The derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)$ at $x=0$ is

The differential coefficient of $\sin^{-1}\left(\frac{1-x}{1+x}\right)$ with respect to $\sqrt{x}$ is:

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} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module