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)

$\frac{2}{1+x^{2}}$, $x \ne0$

Answer 2

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