x=0 આગળ tan ^{-1}left(frac{sqrt{1+x^2}-1}{x}r | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(B) ધારો કે $u = 	an ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}ight)$. $x = 	an 	heta$ લેતા,$u = 	an ^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ight) = 	an ^{-1}\left(\frac{1-\cos 	heta}{\sin 	heta}ight) = 	an ^{-1}\left(	an \frac{	heta}{2}ight) = \frac{	heta}{2} = \frac{1}{2} 	an ^{-1} x$. તેથી,$\frac{du}{dx} = \frac{1}{2(1+x^2)}$. ધારો કે $v = 	an ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}ight)$. $x = \sin \phi$ લેતા,$v = 	an ^{-1}\left(\frac{2 \sin \phi \cos \phi}{1-2 \sin^2 \phi}ight) = 	an ^{-1}(	an 2\phi) = 2\phi = 2 \sin ^{-1} x$. તેથી,$\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$. આપણે $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1}{2(1+x^2)} 	imes \frac{\sqrt{1-x^2}}{2} = \frac{\sqrt{1-x^2}}{4(1+x^2)}$ શોધવાનું છે. $x=0$ આગળ,$\frac{du}{dv} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

$x=0$ આગળ $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ નું $\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)$ ની સાપેક્ષ વિકલન શોધો.

Similar Questions

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

જો $y = \sin^{-1}\left(\frac{\log x^2}{1+(\log x)^2}\right)$ હોય,તો $\left(\frac{dy}{dx}\right)_{x=1} = $

$\tan^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right)$ નો વિકલન સહગુણક શોધો.

જો $y = \sin^{-1}(\sqrt{1 - x^2})$ હોય,તો $dy/dx = $

જો $y=\cos ^{-1}\left(\frac{a^2}{\sqrt{x^4+a^4}}\right)$ હોય,તો $\frac{d y}{d x}$ શું થાય?

Vedclass Test Series

Exam Paper Generator

Online Exam Module