Question

The derivative of ${\cos }^{-1}\frac{x-x^{-1}}{x+x^{-1}}$ is equal to:

• A. $\frac{1}{1+x^2}$
• B. $-\frac{1}{1+x^2}$
• C. $\frac{2}{1+x^2}$
• D. $-\frac{2}{1+x^2}$

Answer 1

Answer: (D)

Let $y={\cos }^{-1}\frac{x-x^{-1}}{x+x^{-1}}$
On differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=-\frac{1}{\sqrt{1-{\left(\frac{x-x^{-1}}{x+x^{-1}}\right)}^2}}\frac{d}{dx}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$
$=-\frac{x+x^{-1}}{\sqrt{{\left(x+x^{-1}\right)}^2-{\left(x-x^{-1}\right)}^2}}$
$\times \frac{\left(x+x^{-1}\right)\left(1+x^{-2}\right)-\left(x-x^{-1}\right)\left(1-x^{-2}\right)}{{\left(x+x^{-1}\right)}^2}$
$=-\frac{1}{\sqrt{4}}(\frac{x+x^{-}1+x^{-}1+x^{-}3-(x-x^{-}1-x^{-}1+x^{-}3)}{(x+x^{-}1)})$
$=-\frac{1}{2}\frac{\left(4x^{-1}\right)}{\left(x+x^{-1}\right)}=-\frac{2}{\left(x^2+1\right)}$
Alternate Solution:
Let $y={\cos }^{-1}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$
$={\cos }^{-1}\left(\frac{x^2-1}{x^2+1}\right)$
Put $x=\cot \theta$
$\therefore y={\cos }^{-1}\left(\frac{{\cot }^2\theta -1}{{\cot }^2\theta +1}\right)$
$={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$
$={\cos }^{-1}\cos 2\theta =2\theta$
$\Rightarrow y=2{\cot }^{-1}x$
On differentiating both sides w. r. t. $x$, we get
$\frac{dy}{dx}=-\frac{2}{1+x^2}$