If u=sin ^{-1}left(frac{2 x}{1+x^2}right) and | vedclass

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

Question

(C) Given $u = \sin^{-1}\left(\frac{2x}{1+x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $u = \sin^{-1}(\sin 2	heta) = 2	heta = 2	an

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given $u = \sin^{-1}\left(\frac{2x}{1+x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $u = \sin^{-1}(\sin 2	heta) = 2	heta = 2	an^{-1}x$.Therefore,$\frac{du}{dx} = \frac{2}{1+x^2}$.Given $v = 	an^{-1}\left(\frac{2x}{1-x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $v = 	an^{-1}(	an 2	heta) = 2	heta = 2	an^{-1}x$.Therefore,$\frac{dv}{dx} = \frac{2}{1+x^2}$.Now,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2/(1+x^2)}{2/(1+x^2)} = 1$.

If $u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$,then $\frac{d u}{d v}$ is

Similar Questions

If $y = \sin^{-1} \left( \frac{25 - x^2}{25 + x^2} \right)$, then find the value of $y'(1)$.

If $y = \tan^{-1} \left( \frac{5x - x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} \right)$,then $\frac{dy}{dx} =$

If $f(x) = \tan^{-1}\left(\frac{1}{\sin^2 x + \sin x + 1}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 3\sin x + 3}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 5\sin x + 7}\right) + \dots$ up to $10$ terms, then $f'(0) = $

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

If $y = \cot^{-1}(\cos 2x)^{1/2}$,then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module