Question

Let $y = \sin^{-1} \left(\frac{2x}{1+x^{2}}\right), 0 < x < 1$ and $0 < y < \frac{\pi}{2},$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{2}{ 1 +x^2}$
• B. $\frac{2x}{ 1 +x^2}$
• C. $- \frac{2}{ 1 +x^2}$
• D. none of these

Answer 1

Answer: (A)

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

Answer 2

Put $x =\tan\theta, \frac{2x}{1+x^{2}} = \frac{2\tan \theta}{1+\tan^{2} \theta} = \sin2 \theta$ $\therefore y = \sin^{-1} \left(\sin2 \theta \right) = 2 \theta = 2 \tan^{-1} x$ $\therefore \frac{dy}{dx} = \frac{2}{1+x^{2}}$