Find the derivative: frac{d}{dx} cos^{-1} lef | vedclass

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

Question

(C) Let $y = \cos^{-1} \left( \frac{x - \frac{1}{x}}{x + \frac{1}{x}} \right) = \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right)$.Substitute $x = \cot

Vedclass · Accepted Answer

$\frac{2}{1 + x^2}$

Vedclass · Answer

(C) Let $y = \cos^{-1} \left( \frac{x - \frac{1}{x}}{x + \frac{1}{x}} ight) = \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} ight)$.Substitute $x = \cot 	heta$,then $x^2 = \cot^2 	heta$.$y = \cos^{-1} \left( \frac{\cot^2 	heta - 1}{\cot^2 	heta + 1} ight) = \cos^{-1} \left( - \frac{1 - \cot^2 	heta}{1 + \cot^2 	heta} ight) = \cos^{-1} (- \cos 2	heta)$.Using the identity $\cos^{-1}(-z) = \pi - \cos^{-1}(z)$,we get $y = \pi - \cos^{-1}(\cos 2	heta) = \pi - 2	heta$.Since $x = \cot 	heta$,$	heta = \cot^{-1} x$.Thus,$y = \pi - 2 \cot^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - 2 \left( - \frac{1}{1 + x^2} ight) = \frac{2}{1 + x^2}$.

Find the derivative: $\frac{d}{dx} \cos^{-1} \left( \frac{x - x^{-1}}{x + x^{-1}} \right)$

Similar Questions

$(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

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

$\operatorname{Tan}^{-1} \left( \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1} \right) + \operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{5}} \right) =$

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module