Question

If \( y = \log(\sec(\tan^{-1}x)) \) for \( x>0 \), then \( \frac{dy}{dx} \) at \( x = 1 \) is:

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

Hint:

For differentiation problems involving inverse trigonometric functions, use standard identities: \[ \sec(\tan^{-1}x) = \sqrt{1 + x^2}, \cosec(\tan^{-1}x) = \frac{\sqrt{1 + x^2}}{x} \]

Answer 1

The Correct Option is C

Define: \[ y = \log(\sec(\tan^{-1}x)) \] Using the identity: \[ \sec(\tan^{-1}x) = \sqrt{1 + x^2} \] Thus, \[ y = \log(\sqrt{1+x^2}) \] Rewriting: \[ y = \frac{1}{2} \log(1 + x^2) \] Differentiating: \[ \frac{dy}{dx} = \frac{1}{2} \times \frac{2x}{1 + x^2} = \frac{x}{1 + x^2} \] Substituting \( x = 1 \): \[ \frac{dy}{dx} = \frac{1}{1 + 1^2} = \frac{1}{2} \] Thus, the correct answer is: \[ \frac{1}{2} \]