Question

If $y={\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$, then $\frac{dy}{dx}$ is equal to

• A. $-\frac{2}{1+x^2}$ of all $|x|<1$
• B. $-\frac{2}{1+x^2}$ of all $|x|>1$
• C. $\frac{2}{1+x^2}$ of all $|x|<1$
• D. None of the above

Answer 1

The correct option is D None of the above

Given that

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

Let $x=\tan \theta$

$\therefore \frac{2x}{1+x^2}=\frac{2\tan \theta }{1+{\tan }^2\theta }$

$=\frac{2\tan \theta }{{\sec }^2\theta }$

$=2\sin \theta \cos \theta =\sin 2\theta$

Hence,

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

$={\cos }^{-1}\left(\sin 2\theta \right)$

$={\cos }^{-1}\left(\cos \right(\frac{\pi }{2}-2\theta \left)\right)$

$=\frac{\pi }{2}-2\theta$

$=\frac{\pi }{2}-2{\tan }^{-1}x$

Hence,

$\frac{dy}{dx}=-2\frac{d}{dx}\left({\tan }^{-1}x\right)$

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

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

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

makes sense if and only if,

$-1\le (\frac{2x}{1+x^2})\le 1$

i.e if

$-(1+x^2)\le 2x\le 1+x^2$

i.e if

$-(1+x^2+2x)\le 0\le 1+x^2-2x$

i.e if

$-(x+1{)}^2\le 0\le (x-1{)}^2$

which holds for all real x.

The domain herein is therefore $(-\infty ,\infty )$.

The required answer is

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

for all real x.

The answer: option D.