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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given that $y = 	an^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} ight)$.Let $x^2 = \cos 2	heta$,then $	heta = \frac{1}{2} \cos^{-1} (x^2)$.Substituting $x^2 = \cos 2	heta$ 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 the identities $1 + \cos 2	heta = 2 \cos^2 	heta$ and $1 - \cos 2	heta = 2 \sin^2 	heta$:$y = 	an^{-1} \left( \frac{\sqrt{2} \cos 	heta + \sqrt{2} \sin 	heta}{\sqrt{2} \cos 	heta - \sqrt{2} \sin 	heta} ight) = 	an^{-1} \left( \frac{\cos 	heta + \sin 	heta}{\cos 	heta - \sin 	heta} ight)$.Dividing numerator and denominator by $\cos 	heta$:$y = 	an^{-1} \left( \frac{1 + 	an 	heta}{1 - 	an 	heta} ight) = 	an^{-1} \left( 	an \left( \frac{\pi}{4} + 	heta ight) ight) = \frac{\pi}{4} + 	heta$.Substituting $	heta = \frac{1}{2} \cos^{-1} (x^2)$:$y = \frac{\pi}{4} + \frac{1}{2} \cos^{-1} (x^2)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 + \frac{1}{2} \left( \frac{-1}{\sqrt{1 - (x^2)^2}} ight) \cdot \frac{d}{dx}(x^2) = \frac{1}{2} \left( \frac{-1}{\sqrt{1 - x^4}} ight) \cdot 2x = \frac{-x}{\sqrt{1 - x^4}}$.

If $y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right)$,where $x^2 \le 1$. Then find $\frac{dy}{dx}$.

Similar Questions

If $y = \operatorname{Tanh}^{-1} \sqrt{\frac{1-x}{1+x}}$,then $\frac{dy}{dx} = $

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

If $y = \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ is equal to

$y = \operatorname{Tan}^{-1}\left(\frac{x}{1+2x^2}\right) + \operatorname{Tan}^{-1}\left(\frac{x}{1+6x^2}\right)$,then $\frac{dy}{dx} = $

Differentiate the following with respect to $x$: $\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module