tan ^{-1}left[frac{x}{1+sqrt{1-x^2}}right] નુ | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{-1}{4}$

Vedclass · Answer

(C) ધારો કે $y = 	an ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}ight)$ અને $u = \sec ^{-1}\left(\frac{1}{2 x^2-1}ight)$.$x = \sin 	heta$ લેતા,$\sqrt{1-x^2} = \cos 	heta$ મળે.$y = 	an ^{-1}\left(\frac{\sin 	heta}{1+\cos 	heta}ight) = 	an ^{-1}\left(\frac{2 \sin(	heta/2) \cos(	heta/2)}{2 \cos^2(	heta/2)}ight) = 	an ^{-1}(	an(	heta/2)) = \frac{	heta}{2}$.હવે,$u = \sec ^{-1}\left(\frac{1}{2 \sin^2 	heta - 1}ight) = \sec ^{-1}\left(\frac{1}{-(1-2 \sin^2 	heta)}ight) = \sec ^{-1}\left(\frac{1}{-\cos 2 	heta}ight) = \sec ^{-1}(-\sec 2 	heta) = \sec ^{-1}(\sec(\pi - 2 	heta)) = \pi - 2 	heta$.$x = \sin 	heta$ હોવાથી,$	heta = \sin^{-1} x$.તેથી,$y = \frac{1}{2} \sin^{-1} x$ અને $u = \pi - 2 \sin^{-1} x$.હવે $\frac{dy}{dx} = \frac{1}{2 \sqrt{1-x^2}}$ અને $\frac{du}{dx} = \frac{-2}{\sqrt{1-x^2}}$.તેથી,$\frac{dy}{du} = \frac{dy/dx}{du/dx} = \frac{1/(2 \sqrt{1-x^2})}{-2/\sqrt{1-x^2}} = -\frac{1}{4}$.

$\tan ^{-1}\left[\frac{x}{1+\sqrt{1-x^2}}\right]$ નું $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ ની સાપેક્ષમાં વિકલન શું થાય?

Similar Questions

જો $y = \tan^{-1}(\sec x + \tan x)$ હોય,તો $\frac{dy}{dx} = $

${\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ નું ${\cot ^{ - 1}}\left( {\frac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right)$ ની સાપેક્ષમાં વિકલન શું થાય?

જો $f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1+9^x}\right)$ હોય,તો $f^{\prime}\left(\frac{1}{2}\right)$ ની કિંમત શોધો.

જો $0 < |x| < 1$ માટે $f(x) = \operatorname{Tan}^{-1} \left[ \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right]$ હોય,તો $f'(x) =$

જો $y=\tan ^{-1}\left\{\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right\}$ હોય,તો $\frac{d y}{d x}$ શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module