The derivative of tan^{-1} sqrt{frac{1-x}{1+x | vedclass

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

Question

(B) Let $u = 	an^{-1} \sqrt{\frac{1-x}{1+x}}$ and $v = \sin^{-1} x$.Substitute $x = \cos 	heta$,where $	heta = \cos^{-1} x$.Then $\sqrt{\frac{1-x}{

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(B) Let $u = 	an^{-1} \sqrt{\frac{1-x}{1+x}}$ and $v = \sin^{-1} x$.Substitute $x = \cos 	heta$,where $	heta = \cos^{-1} x$.Then $\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{2\sin^2(	heta/2)}{2\cos^2(	heta/2)}} = 	an(	heta/2)$.So,$u = 	an^{-1}(	an(	heta/2)) = \frac{	heta}{2} = \frac{1}{2} \cos^{-1} x$.Since $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$,we have $u = \frac{1}{2} (\frac{\pi}{2} - v) = \frac{\pi}{4} - \frac{1}{2} v$.Now,differentiate $u$ with respect to $v$:$\frac{du}{dv} = \frac{d}{dv} (\frac{\pi}{4} - \frac{1}{2} v) = -\frac{1}{2}$.

The derivative of $\tan^{-1} \sqrt{\frac{1-x}{1+x}}$ with respect to $\sin^{-1}x$ is -

Similar Questions

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \right] = $

The derivative of $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$ with respect to $\sin ^{-1}\left(3 x-4 x^3\right)$ 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

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{1+6x^2} \right) \right) = $ . . . . . .

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module