Question

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

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

Answer 1

The correct option is A $-\frac{2}{\sqrt{(1-x^2)}}.$

Let $y={\cos }^{-1}\frac{x-x^{-1}}{x+x^{-1}}$

$={\cos }^{-1}\frac{x^2-1}{x^2+1}$

$=\pi -{\cos }^{-1}\frac{1-x^2}{1+x^2}$
Put $x=\tan \theta$
$\Rightarrow y=\pi -{\cos }^{-1}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\pi -{\cos }^{-1}\cos 2\theta$

$=\pi -2\theta =\pi -2{\tan }^{-1}x$
$\therefore \frac{dy}{dx}=-\frac{2}{1+x^2}$