The differential coefficient of tan^{-1}left( | vedclass

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

Question

(C) Let $y = 	an^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} ight)$.Put $x = \cos 2	heta$,which implies $	heta = \frac{1}{

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{1-x^2}}$

Vedclass · Answer

(C) Let $y = 	an^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} ight)$.Put $x = \cos 2	heta$,which implies $	heta = \frac{1}{2}\cos^{-1}x$.Substituting $x$ into the expression:$y = 	an^{-1}\left( \frac{\sqrt{1+\cos 2	heta} - \sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta} + \sqrt{1-\cos 2	heta}} ight)$Using trigonometric identities $1+\cos 2	heta = 2\cos^2	heta$ and $1-\cos 2	heta = 2\sin^2	heta$:$y = 	an^{-1}\left( \frac{\sqrt{2\cos^2	heta} - \sqrt{2\sin^2	heta}}{\sqrt{2\cos^2	heta} + \sqrt{2\sin^2	heta}} ight) = 	an^{-1}\left( \frac{\cos	heta - \sin	heta}{\cos	heta + \sin	heta} ight)$Dividing numerator and denominator by $\cos	heta$:$y = 	an^{-1}\left( \frac{1 - 	an	heta}{1 + 	an	heta} ight) = 	an^{-1}(	an(\pi/4 - 	heta))$Thus,$y = \frac{\pi}{4} - 	heta = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} ight) = \frac{1}{2\sqrt{1-x^2}}$.

The differential coefficient of $\tan^{-1}\left( \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right)$ is

Similar Questions

$\lim _{x}$ ${\rightarrow \frac{\pi}{2}} \left( \frac{\int_{x^3}^{(\pi / 2)^3} (\sin (2 t^{1 / 3}) + \cos (t^{1 / 3})) dt}{(x - \frac{\pi}{2})^2} \right)$ is equal to:

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

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

Let $f : R \rightarrow R$ be a differentiable function and $f(1) = 4$. Then the value of $\lim_{x \rightarrow 1} \int_{4}^{f(x)} \frac{2t \, dt}{x - 1}$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module