If y = sin^{-1} left[ frac{sqrt{1+x} + sqrt{1 | vedclass

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

Question

(A) Let $x = \cos(2	heta)$,so $2	heta = \cos^{-1}(x)$ and $	heta = \frac{1}{2} \cos^{-1}(x)$.Then $\sqrt{1+x} = \sqrt{1+\cos(2	heta)} = \sqrt{2\co

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $x = \cos(2	heta)$,so $2	heta = \cos^{-1}(x)$ and $	heta = \frac{1}{2} \cos^{-1}(x)$.Then $\sqrt{1+x} = \sqrt{1+\cos(2	heta)} = \sqrt{2\cos^2(	heta)} = \sqrt{2} \cos(	heta)$ and $\sqrt{1-x} = \sqrt{1-\cos(2	heta)} = \sqrt{2\sin^2(	heta)} = \sqrt{2} \sin(	heta)$.Substituting these into the expression for $y$:$y = \sin^{-1} \left[ \frac{\sqrt{2}\cos(	heta) + \sqrt{2}\sin(	heta)}{2} ight] = \sin^{-1} \left[ \frac{1}{\sqrt{2}} \cos(	heta) + \frac{1}{\sqrt{2}} \sin(	heta) ight]$.Using the identity $\sin(\frac{\pi}{4} + 	heta) = \sin(\frac{\pi}{4}) \cos(	heta) + \cos(\frac{\pi}{4}) \sin(	heta) = \frac{1}{\sqrt{2}} \cos(	heta) + \frac{1}{\sqrt{2}} \sin(	heta)$.So,$y = \sin^{-1} [\sin(\frac{\pi}{4} + 	heta)] = \frac{\pi}{4} + 	heta = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 + \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} ight) = -\frac{1}{2\sqrt{1-x^2}}$.

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

Similar Questions

The differential coefficient of ${\tan ^{ - 1}}\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} $ with respect to ${\cos ^{ - 1}}({x^2})$ is

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

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

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module