Find frac{dy}{dx} if y = sin^{-1}left(frac{1- | vedclass

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

Question

(A) Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$. Since $0 < x < 1$,we have $0 < 	heta < \frac{\pi}{4}$.The expression becomes $y = \sin^{-1}\le

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$. Since $0 The expression becomes $y = \sin^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight)$.Using the trigonometric identity $\cos 2	heta = \frac{1-	an^2 	heta}{1+	an^2 	heta}$,we get $y = \sin^{-1}(\cos 2	heta)$.Since $\cos 2	heta = \sin\left(\frac{\pi}{2} - 2	hetaight)$,we have $y = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2	hetaight)ight)$.Given $0 Thus,$y = \frac{\pi}{2} - 2	heta = \frac{\pi}{2} - 2	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}ight) - 2\frac{d}{dx}(	an^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}ight) = \frac{-2}{1+x^2}$.

Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

Similar Questions

$\sec ^2(\tan ^{-1} 2) + \operatorname{cosec}^2(\cot ^{-1} 3)$ is equal to

Solve $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2},$ then $x$ is equal to

If $\sinh (2 \tanh ^{-1} x) = \frac{11}{60}$,then $x =$

Assume that $3.13 \leq \pi \leq 3.15$. The integer closest to the value of $\sin ^{-1}(\sin 1 \cos 4+\cos 1 \sin 4)$,where $1$ and $4$ appearing in $\sin$ and $\cos$ are given in radians,is

If $u = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)$ and $v = 2 \tan^{-1} x$,then $\frac{du}{dv}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module