If y = cot^{-1}(cos 2x)^{1/2},then the value | vedclass

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

Question

(A) Given $y = \cot^{-1}(\sqrt{\cos 2x})$.Using the chain rule,$\frac{dy}{dx} = -\frac{1}{1 + (\sqrt{\cos 2x})^2} \cdot \frac{d}{dx}(\sqrt{\cos 2x})$.

Vedclass · Accepted Answer

$\left(\frac{2}{3}\right)^{1/2}$

Vedclass · Answer

(A) Given $y = \cot^{-1}(\sqrt{\cos 2x})$.Using the chain rule,$\frac{dy}{dx} = -\frac{1}{1 + (\sqrt{\cos 2x})^2} \cdot \frac{d}{dx}(\sqrt{\cos 2x})$.$\frac{dy}{dx} = -\frac{1}{1 + \cos 2x} \cdot \frac{1}{2\sqrt{\cos 2x}} \cdot (-\sin 2x \cdot 2)$.$\frac{dy}{dx} = \frac{\sin 2x}{(1 + \cos 2x)\sqrt{\cos 2x}}$.Using the identities $\sin 2x = 2\sin x \cos x$ and $1 + \cos 2x = 2\cos^2 x$,we get:$\frac{dy}{dx} = \frac{2\sin x \cos x}{2\cos^2 x \sqrt{\cos 2x}} = \frac{	an x}{\sqrt{\cos 2x}}$.At $x = \frac{\pi}{6}$,$	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$ and $\cos 2(\frac{\pi}{6}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$.Substituting these values: $\frac{dy}{dx} = \frac{1/\sqrt{3}}{\sqrt{1/2}} = \frac{1}{\sqrt{3}} \cdot \sqrt{2} = \sqrt{\frac{2}{3}} = \left(\frac{2}{3}ight)^{1/2}$.

If $y = \cot^{-1}(\cos 2x)^{1/2}$,then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is:

Similar Questions

Find the value of $\frac{d}{dx} \left[ \tan^{-1} \sqrt{\frac{1 - \cos x}{1 + \cos x}} \right]$.

If $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$,then $\frac{dy}{dx} = $

If $y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$,then $\frac{dy}{dx} = \ldots$

Let $f: R \rightarrow R$ be a continuous function. If $px+my+n=0$ is a tangent drawn to the curve $y=f(x)$ at $x=\alpha$,then at $x=0$,$\frac{d}{d x}\left(f\left(\alpha e^{2 x}\right)\right)=$

If $u=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ and $v=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$,then $\frac{d u}{d v}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module