If y = sin^{-1} left( frac{2x}{1 + x^2} right | vedclass

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

Question

(C) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$.The expression becomes $y = \sin^{-1} \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight) +

Vedclass · Accepted Answer

$\frac{4}{1 + x^2}$

Vedclass · Answer

(C) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$.The expression becomes $y = \sin^{-1} \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight) + \sec^{-1} \left( \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta} ight)$.Using trigonometric identities,$\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$ and $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$.Thus,$y = \sin^{-1} (\sin 2	heta) + \sec^{-1} \left( \frac{1}{\cos 2	heta} ight) = \sin^{-1} (\sin 2	heta) + \sec^{-1} (\sec 2	heta)$.Assuming the principal value branch,$y = 2	heta + 2	heta = 4	heta$.Substituting back $	heta = 	an^{-1} x$,we get $y = 4 	an^{-1} x$.Differentiating with respect to $x$,$\frac{dy}{dx} = 4 \cdot \frac{d}{dx} (	an^{-1} x) = \frac{4}{1 + x^2}$.

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

Similar Questions

If $y = \tan^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$, then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is

Derivative of $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ with respect to $\tan ^{-1} x$ for $-1 < x < 1$ is:

Find the value of $\frac{d}{dx} [\sin^2 \{ \cot^{-1} \sqrt{\frac{1-x}{1+x}} \}]$.

Differentiation of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to $\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right)$ is

The derivative of $\sin ^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)$ with respect to $\cos ^{-1} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module