Question

If \(y = \sec (\tan^{-1} x)\), then \(\frac{dy}{dx}\) at \(x = \sqrt{3}\) is equal to

• A. \(\frac{2\sqrt{3}}{3}\)
• B. \(\frac{1}{2}\)
• C. 2
• D. \(\frac{\sqrt{3}}{2}\)
• E. \(2\sqrt{3}\)

Hint:

\(\sec(\tan^{-1} x) = \sqrt{1 + x^2}\) is a useful identity.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Let \(\theta = \tan^{-1} x\). Then \(\sec(\tan^{-1} x) = \sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + x^2}\).

Step 2: Detailed Explanation:
\(y = \sqrt{1 + x^2}\)
\(\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^2}} \cdot 2x = \frac{x}{\sqrt{1 + x^2}}\)
At \(x = \sqrt{3}\): \(\frac{dy}{dx} = \frac{\sqrt{3}}{\sqrt{1 + 3}} = \frac{\sqrt{3}}{2}\)

Step 3: Final Answer:
\(\frac{dy}{dx} = \frac{\sqrt{3}}{2}\).