Question

If y = sin^–1$\frac{1}{\sqrt{1 + x^{2}}}$+ tan^–1$\left( \frac{\sqrt{1 + x^{2}} - 1}{x} \right)$ , then the $\frac{dy}{dx}$

equals –

• A. $\frac{1}{2(1 + x^{2})}$
• B. – $\frac{1}{2(1 + x^{2})}$
• C. 0
• D. None of these

Answer 1

Answer: (B)

– $\frac{1}{2(1 + x^{2})}$

Answer 2

Put x = tan θ

y = sin^–1 $\frac{1}{\sec\theta}$+ tan^–1 $\left( \frac{\sec\theta - 1}{\tan\theta} \right)$

⇒ y = $\left( \frac{\pi}{2} - \theta \right)$+ tan^–1 $\left( \frac{1 - \cos\theta}{\sin\theta} \right)$

⇒ y = $\frac{\pi}{2}$ –q + $\frac{\theta}{2}$