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 $f(x) = \frac{1}{1 + \frac{1}{x}}$ and $g(x) = \frac{1}{1 + \frac{1}{f(x)}}$, then $g^{\prime}(2)$ is equal to

Find the derivative: $\frac{d}{dx} \left[ \frac{e^{ax}}{\sin(bx + c)} \right]$

If $y = \tan^{-1}\left( \frac{\sqrt{x} - x}{1 + x^{3/2}} \right)$,then $y'(1)$ is

The derivative of $\sqrt{\sqrt{x} + 1}$ is

Differentiate the function with respect to $x$: $2 \sqrt{\cot \left(x^{2}\right)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module