If y = tan^{-1} left{ frac{sqrt{1 + x^2} - sq | vedclass

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

Question

(B) Let $x^2 = \cos 	heta$, so $	heta = \cos^{-1}(x^2)$.Then $y = 	an^{-1} \left\{ \frac{\sqrt{1 + \cos 	heta} - \sqrt{1 - \cos 	heta}}{\sqrt{1 +

Vedclass · Accepted Answer

$\frac{x}{\sqrt{1 - x^4}}$

Vedclass · Answer

(B) Let $x^2 = \cos 	heta$, so $	heta = \cos^{-1}(x^2)$.Then $y = 	an^{-1} \left\{ \frac{\sqrt{1 + \cos 	heta} - \sqrt{1 - \cos 	heta}}{\sqrt{1 + \cos 	heta} + \sqrt{1 - \cos 	heta}} ight\}$.Using half-angle formulas $\sqrt{1 + \cos 	heta} = \sqrt{2} \cos(	heta/2)$ and $\sqrt{1 - \cos 	heta} = \sqrt{2} \sin(	heta/2)$, we get:$y = 	an^{-1} \left\{ \frac{\sqrt{2} \cos(	heta/2) - \sqrt{2} \sin(	heta/2)}{\sqrt{2} \cos(	heta/2) + \sqrt{2} \sin(	heta/2)} ight\} = 	an^{-1} \left\{ \frac{1 - 	an(	heta/2)}{1 + 	an(	heta/2)} ight\}$.$y = 	an^{-1} \left\{ 	an(\frac{\pi}{4} - \frac{	heta}{2}) ight\} = \frac{\pi}{4} - \frac{	heta}{2} = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}(x^2)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - \frac{1}{2} \left( -\frac{1}{\sqrt{1 - (x^2)^2}} \cdot 2x ight) = \frac{x}{\sqrt{1 - x^4}}$.

If $y = \tan^{-1} \left\{ \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right\}$, where $|x| < 1$, then $\frac{dy}{dx}$ is equal to

Similar Questions

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

If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$, then the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is

Let $f: R \rightarrow R$ be a continuous function. If $px+my+n=0$ is a tangent drawn to the curve $y=f(x)$ at $x=\alpha$,then at $x=0$,$\frac{d}{d x}\left(f\left(\alpha e^{2 x}\right)\right)=$

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) = $

If $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module