frac{d}{d x}left[cos ^2left(cot ^{-1} sqrt{fr | vedclass

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

Question

(D) Let $	heta = \cot ^{-1} \sqrt{\frac{2+x}{2-x}}$. Then $\cot 	heta = \sqrt{\frac{2+x}{2-x}}$.From the right-angled triangle with base $\sqrt{2+x}

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(D) Let $	heta = \cot ^{-1} \sqrt{\frac{2+x}{2-x}}$. Then $\cot 	heta = \sqrt{\frac{2+x}{2-x}}$.From the right-angled triangle with base $\sqrt{2+x}$ and perpendicular $\sqrt{2-x}$,the hypotenuse is $\sqrt{(\sqrt{2+x})^2 + (\sqrt{2-x})^2} = \sqrt{2+x+2-x} = \sqrt{4} = 2$.Thus,$\cos 	heta = \frac{	ext{base}}{	ext{hypotenuse}} = \frac{\sqrt{2+x}}{2}$.Now,the expression becomes $\frac{d}{d x}\left[\cos ^2 	hetaight] = \frac{d}{d x}\left[\left(\frac{\sqrt{2+x}}{2}ight)^2ight]$.$= \frac{d}{d x}\left(\frac{2+x}{4}ight) = \frac{d}{d x}\left(\frac{1}{2} + \frac{x}{4}ight) = 0 + \frac{1}{4} = \frac{1}{4}$.

$\frac{d}{d x}\left[\cos ^2\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is

Similar Questions

$\frac{d}{dx} \log |x| = ......, (x \ne 0)$

If $f(x) = x \tan^{-1} x$,then $\lim_{x \rightarrow 1} \frac{f(x) - f(1)}{x - 1}$ is equal to

$\frac{d}{dx}\{ \log (\sec x + \tan x)\} = $

For the function $f(x) = x^2 - 6x + 8$ where $2 \le x \le 4$,the value of $x$ for which $f'(x)$ vanishes is:

Find the derivative of the function: $(x^{2}+1) \cos x$. (Assume that $a, b, c, d, p, q, r, s$ are fixed non-zero constants and $m, n$ are integers.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module