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 $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$,then the values of $x$ are

The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$ is

Prove $3 \sin ^{-1} x = \sin ^{-1}(3 x - 4 x^{3})$,where $x \in [-\frac{1}{2}, \frac{1}{2}]$.

$\tan^{-1} \left[ \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right] = $

Prove that $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{31}{17}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module