The derivative of operatorname{Sec}^{-1}left( | vedclass

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

Question

(D) Let $u = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ and $v = \sqrt{1-x^2}$.Using the identity $\operatorname{Sec}^{-1}(\frac{1}{z}) = \

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let $u = \operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}ight)$ and $v = \sqrt{1-x^2}$.Using the identity $\operatorname{Sec}^{-1}(\frac{1}{z}) = \cos^{-1}(z)$,we have $u = \cos^{-1}(2x^2-1)$.Let $x = \cos 	heta$,then $u = \cos^{-1}(2\cos^2 	heta - 1) = \cos^{-1}(\cos 2	heta) = 2	heta = 2\cos^{-1}x$.Thus,$\frac{du}{dx} = 2 	imes (-\frac{1}{\sqrt{1-x^2}}) = -\frac{2}{\sqrt{1-x^2}}$.Now,$v = \sqrt{1-x^2}$,so $\frac{dv}{dx} = \frac{1}{2\sqrt{1-x^2}} 	imes (-2x) = -\frac{x}{\sqrt{1-x^2}}$.We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-2/\sqrt{1-x^2}}{-x/\sqrt{1-x^2}} = \frac{2}{x}$.At $x = \frac{1}{2}$,$\frac{du}{dv} = \frac{2}{1/2} = 4$.

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

Similar Questions

If $y = \operatorname{Tan}^{-1}\left(\frac{2x}{1-x^2}\right)$ where $|x| < 1$,then find the value of $\left(\frac{dy}{dx}\right)$ at $x = \frac{1}{2}$.

If $f(x) = \operatorname{Tan}^{-1} \left[ \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right]$ for $0 < |x| < 1$,then $f'(x) =$

If $y = \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ is equal to

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module