The derivative of the function cot ^{-1}[cos | vedclass

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

Question

(B) Let $f(x) = \cot^{-1}(\sqrt{\cos 2x})$.Using the chain rule,the derivative is $f'(x) = \frac{-1}{1 + (\sqrt{\cos 2x})^2} 	imes \frac{d}{dx}(\sqrt

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $f(x) = \cot^{-1}(\sqrt{\cos 2x})$.Using the chain rule,the derivative is $f'(x) = \frac{-1}{1 + (\sqrt{\cos 2x})^2} 	imes \frac{d}{dx}(\sqrt{\cos 2x})$.$f'(x) = \frac{-1}{1 + \cos 2x} 	imes \frac{1}{2\sqrt{\cos 2x}} 	imes (-\sin 2x 	imes 2)$.$f'(x) = \frac{\sin 2x}{(1 + \cos 2x)\sqrt{\cos 2x}}$.At $x = \frac{\pi}{6}$,$2x = \frac{\pi}{3}$.$f'\left(\frac{\pi}{6}ight) = \frac{\sin(\pi/3)}{(1 + \cos(\pi/3))\sqrt{\cos(\pi/3)}}$.$f'\left(\frac{\pi}{6}ight) = \frac{\sqrt{3}/2}{(1 + 1/2)\sqrt{1/2}} = \frac{\sqrt{3}/2}{(3/2)(1/\sqrt{2})} = \frac{\sqrt{3}}{2} 	imes \frac{2}{3} 	imes \sqrt{2} = \frac{\sqrt{6}}{3} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}} = \left(\frac{2}{3}ight)^{1/2}$.

The derivative of the function $\cot ^{-1}[\cos 2 x]^{1 / 2}$ at $x=\pi / 6$ is

Similar Questions

If $f: R \rightarrow R$ is an even function having derivatives of all orders, then which of the following is an odd function?

If $f(x) = \sqrt{\cos^{-1} \sqrt{1-x^2}}$, then $f^{\prime}\left(\frac{1}{2}\right) = $

If $f(x) = \sum_{p=1}^7 p^2 \sin^{-1}\left(\frac{4}{5} \sin(px) - \frac{3}{5} \cos(px)\right)$, then the value of $\frac{df}{dx}$ at $x = 1$ is (Given that $\sin^{-1}(\sin x) = x$)

If $y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$,then find $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module