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

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

Question

(B) Let $u = 	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$. Put $x = 	an 	heta$,then $u = 	an ^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ig

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(B) Let $u = 	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$. Put $x = 	an 	heta$,then $u = 	an ^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ight) = 	an ^{-1}\left(\frac{1-\cos 	heta}{\sin 	heta}ight) = 	an ^{-1}\left(	an \frac{	heta}{2}ight) = \frac{	heta}{2} = \frac{1}{2} 	an ^{-1} x$. Thus,$\frac{du}{dx} = \frac{1}{2(1+x^2)}$. Let $v = 	an ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}ight)$. Put $x = \sin \phi$,then $v = 	an ^{-1}\left(\frac{2 \sin \phi \cos \phi}{1-2 \sin^2 \phi}ight) = 	an ^{-1}(	an 2\phi) = 2\phi = 2 \sin ^{-1} x$. Thus,$\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1}{2(1+x^2)} 	imes \frac{\sqrt{1-x^2}}{2} = \frac{\sqrt{1-x^2}}{4(1+x^2)}$. At $x=0$,$\frac{du}{dv} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

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

Similar Questions

If $a > b > 0$ and $x$ is acute,then $\frac{d}{dx} \left[ \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} \right) \right] = $

If $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left[ \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} \right]$,then $\frac{dy}{dx} = $

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

If $y = \sin^{-1} \left[ \frac{\sqrt{1+x} + \sqrt{1-x}}{2} \right]$,then $\frac{dy}{dx} = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module