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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $y = 	an ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}ight)$.Substitute $x = \cos 2	heta$,which implies $	heta = \frac{1}{2} \cos ^{-1} x$.Then $y = 	an ^{-1}\left(\frac{\sqrt{1+\cos 2	heta}-\sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta}+\sqrt{1-\cos 2	heta}}ight)$.Using trigonometric identities $1+\cos 2	heta = 2\cos^2	heta$ and $1-\cos 2	heta = 2\sin^2	heta$,we get:$y = 	an ^{-1}\left(\frac{\sqrt{2}\cos	heta - \sqrt{2}\sin	heta}{\sqrt{2}\cos	heta + \sqrt{2}\sin	heta}ight) = 	an ^{-1}\left(\frac{1-	an	heta}{1+	an	heta}ight)$.Using the formula $	an(\frac{\pi}{4} - 	heta) = \frac{1-	an	heta}{1+	an	heta}$,we have:$y = 	an ^{-1}	an(\frac{\pi}{4} - 	heta) = \frac{\pi}{4} - 	heta = \frac{\pi}{4} - \frac{1}{2} \cos ^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} 	imes \left(-\frac{1}{\sqrt{1-x^2}}ight) = \frac{1}{2 \sqrt{1-x^2}}$.

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

Similar Questions

If $f:R \to R$ is a differentiable function and $f(2) = 6$,then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{2t \, dt}{x - 2}$ is

If $y=\sin ^{-1}\left[x \sqrt{1-x^2}-\sqrt{x} \sqrt{1-x}\right]$ and $0 < x < 1$,then $\frac{d y}{d x}$ is equal to

If $\sqrt{1 - x^6} + \sqrt{1 - y^6} = a^3(x^3 - y^3)$,then $\frac{dy}{dx} = $

Find the value of $\frac{d}{dx} \left[ \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right]$.

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) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module