Question

$\frac{d}{dx}\left\{cose{c}^{-1}\left(\frac{1+x^2}{2x}\right)\right\}$ is equal to

• A. $\frac{2}{1+x^2}$, $x\ne 0$
• B. $\frac{2\left(1+x\right)}{1+x^2}$, $x\ne 0$
• C. $\frac{2\left(1-x^2\right)}{\left(1+x^2\right)|1-x^2|}$, $x\ne \pm 1$, $0$
• D. None of these

Answer 1

Answer: (A)

let $y=cose{c}^{-1}\left(\frac{1+x^2}{2x}\right)$
$y=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$
put $x=tan\theta ,\theta =ta{n}^{-1}x$
$=si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$
$=si{n}^{-1}(sin2\theta )$
$y=2\theta =2ta{n}^{-1}x$
$\frac{dy}{dx}=\frac{2}{1+x^2}$
$\therefore \frac{2}{1+x^2}$, $x\ne 0iscorrect$