Question

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

• A. $\frac{2}{1+x^{2}}$, $x \ne0$
• B. $\frac{2\left(1+x\right)}{1+x^{2}}$, $x \ne0$
• C. $\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)\left|1-x^{2}\right|}$, $x \ne\pm1$, $0$
• D. None of these

Answer 1

Answer: (A)

let $ y = cosec^{-1} \left(\frac{1+x^{2}}{2x}\right)$
$y = sin^{-1}\left(\frac{2x}{1+x^{2}}\right)$
put $x = tan \theta, \,\,\, \theta = tan^{-1}x$
$= sin^{-1}\left(\frac{2 tan \theta}{1+tan^{2} \theta}\right)$
$= sin^{-1} (sin 2\theta)$
$y = 2\theta = 2tan^{-1}x$
$\frac{dy}{dx} = \frac{2}{1+x^2}$
$\therefore \frac{2}{1+x^{2}}$, $x \ne0 is correct $