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

If $y=\sec ^{-1}\left(\frac{x+x^{-1}}{x-x^{-1}}\right)$,then $\frac{d y}{d x}=$

If $y = \tan^{-1}\left(\frac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \dots$ up to $n$ terms,then $\frac{dy}{dx}$ is equal to

If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ has the value

If $y = \tan^{-1}(\sec x + \tan x)$,then $\frac{dy}{dx} = $

If $\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3(x^3 - y^3)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module