The derivative of sec^{-1}left( frac{1}{2x^2 | vedclass

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

Question

(A) Let $u = \sec^{-1}\left( \frac{1}{2x^2 - 1} ight) = \cos^{-1}(2x^2 - 1)$.Substitute $x = \cos 	heta$,then $u = \cos^{-1}(2\cos^2 	heta - 1) =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let $u = \sec^{-1}\left( \frac{1}{2x^2 - 1} ight) = \cos^{-1}(2x^2 - 1)$.Substitute $x = \cos 	heta$,then $u = \cos^{-1}(2\cos^2 	heta - 1) = \cos^{-1}(\cos 2	heta) = 2	heta = 2\cos^{-1} x$.Thus,$\frac{du}{dx} = 2 	imes \left( -\frac{1}{\sqrt{1 - x^2}} ight) = -\frac{2}{\sqrt{1 - x^2}}$.Let $v = \sqrt{1 - x^2}$.Then $\frac{dv}{dx} = \frac{1}{2\sqrt{1 - x^2}} 	imes (-2x) = -\frac{x}{\sqrt{1 - x^2}}$.Now,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-2/\sqrt{1 - x^2}}{-x/\sqrt{1 - x^2}} = \frac{2}{x}$.At $x = \frac{1}{2}$,$\frac{du}{dv} = \frac{2}{1/2} = 4$.

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

Similar Questions

If $y = \tan^{-1}\left(\frac{12x - 64x^3}{1 - 48x^2}\right)$,then $\frac{dy}{dx} = $

For $-1 < x < 1$,if $f(x) = \cos^2 \left( \tan^{-1} \sqrt{\frac{1-x}{1+x}} \right)$,then $f'(x) =$

If $y = \tan^{-1}\left( \frac{a\cos x - b\sin x}{b\cos x + a\sin x} \right)$,then $\frac{dy}{dx} = $

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

$\begin{aligned} & \text{If } y = \tan^{-1} \left\{ \frac{x}{1 + \sqrt{1 - x^2}} \right\} \\ & + \sin \left\{ 2 \tan^{-1} \sqrt{\frac{1 - x}{1 + x}} \right\} \text{, then } \frac{dy}{dx} = \end{aligned}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module