Question

If $$ y = \sin^{-1} \left( \frac{2x}{1+x^2} \right) + \sec^{-1} \left( \frac{1+x^2}{1-x^2} \right) $$ then the value of $$ \frac{dy}{dx} $$ at $$ x = \sqrt{3} $$ is

• A. \(1\)
• B. \(1/2\)
• C. \(0\)
• D. \(1/4\)
• E. None of these

Hint:

In differentiation questions, always check first whether the function is:
• explicit,
• implicit,
• or parametric. e} Only after knowing the full equation can \(\frac{dy}{dx}\) be evaluated at a specific value of \(x\).

Answer 1

The Correct Option is A

$$ y = \sin^{-1} \left( \frac{2x}{1+x^2} \right) + \sec^{-1} \left( \frac{1+x^2}{1-x^2} \right) $$

Put $ x = \tan \theta \Rightarrow \theta = \tan^{-1} x $

$$ \begin{aligned} \therefore \quad y &= \sin^{-1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) + \sec^{-1} \left( \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} \right) \\ &= \sin^{-1}(\sin 2\theta) + \sec^{-1}(\sec 2\theta) \\ &= 2\theta + 2\theta \\ &= 4\theta \\ &= 4 \tan^{-1} x \end{aligned} $$

$$ \therefore \quad \frac{dy}{dx} = \frac{4}{1 + x^2} $$

At $ x = \sqrt{3} $

$$ \frac{dy}{dx} = \frac{4}{1 + (\sqrt{3})^2} = \frac{4}{1 + 3} = 1 $$