Find frac{dy}{dx} for y = cos^{-1}left(frac{2 | vedclass

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

Question

(A) Given $y = \cos^{-1}\left(\frac{2x}{1+x^2}ight)$.Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Since $-1 < x < 1$,we have $-\frac{\pi}{4}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $y = \cos^{-1}\left(\frac{2x}{1+x^2}ight)$.Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Since $-1 Then $\frac{2x}{1+x^2} = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \sin(2	heta)$.So,$y = \cos^{-1}(\sin(2	heta)) = \cos^{-1}(\cos(\frac{\pi}{2} - 2	heta))$.Since $-\frac{\pi}{4} Thus,$y = \frac{\pi}{2} - 2	heta = \frac{\pi}{2} - 2 	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} = \frac{-2}{1+x^2}$.

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

Similar Questions

If the sum of all the solutions of $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}$ for $-1 < x < 1$ and $x \neq 0$ is $\alpha-\frac{4}{\sqrt{3}}$,then $\alpha$ is equal to $..........$.

If $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$,then $\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = $

Find the value of $\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)$.

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

If $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$,then $p^2 + q^2 + r^2 + 2pqr = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module