Question

If $y = \tan^{-1} \frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}} ,$ then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{1 -x^2}$
• B. $\frac{1}{\sqrt{1 -x^2}}$
• C. $\frac{1}{1 + x^2}$
• D. $\frac{1}{\sqrt{1 + x^2}}$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{1 -x^2}}$

Answer 2

Put $x = \cos\theta$ $y = \tan^{-1} \frac{\cos\theta - \sin \theta}{\cos\theta + \sin\theta} =\tan^{-1} \frac{1-\tan\theta}{1+\tan\theta}$ $=\tan^{-1} \tan\left(\frac{\pi}{4} - \theta\right)$ $= \frac{\pi}{4} - \theta = \frac{\pi}{4} -\cos^{-1} x$ $\therefore \frac{dy}{dx} = 0 - \left(\frac{1}{\sqrt{1-x^{2}}} \right) = \frac{1}{\sqrt{1-x^{2}}}$