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
$\frac{d y}{d x}=-\frac{1}{\sqrt{1-\left(\frac{x-x^{-1}}{x+x^{-1}}\right)^{2}}} \frac{d}{d x}\left(\frac{x-x^{-1}}{x+x^{-1}}\right)$
$=-\frac{x+x^{-1}}{\sqrt{\left(x+x^{-1}\right)^{2}-\left(x-x^{-1}\right)^{2}}}$
$\times \frac{\left(x+x^{-1}\right)\left(1+x^{-2}\right)-\left(x-x^{-1}\right)\left(1-x^{-2}\right)}{\left(x+x^{-1}\right)^{2}}$
$ = - \frac{1}{\sqrt{4}} (\frac{x+x^-1+x^-1+x^-3-(x-x^-1-x^-1+x^-3)}{(x+x^-1)})$
$=-\frac{1}{2} \frac{\left(4 x^{-1}\right)}{\left(x+x^{-1}\right)}=-\frac{2}{\left(x^{2}+1\right)}$
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
$\frac{d y}{d x}=-\frac{2}{1+x^{2}}$