Question

If \( y = \tan(\log x) \), then \( \frac{d^2y}{dx^2} \) is given by: 

• A. \( \frac{-\sec^2(\log x) [1 + 2\tan x]}{x^2} \)
• B. \( \frac{\sec^2(\log x) [1 + \tan(\log x)]}{x^2} \)
• C. \( \frac{\sec(\log x) [2\tan(\log x) - 1]}{x^2} \)
• D. \( \frac{\sec^2(\log x) [2\tan(\log x) - 1]}{x^2} \)

Hint:

When differentiating functions involving logarithmic arguments, use the chain rule carefully. The second derivative often requires the quotient rule or further simplifications.

Answer 1

The Correct Option is D

Step 1: Differentiate \( y = \tan(\log x) \)
We are given: \[ y = \tan(\log x). \] Differentiating both sides with respect to \( x \): \[ \frac{dy}{dx} = \sec^2(\log x) \cdot \frac{d}{dx} (\log x). \] Since \( \frac{d}{dx} (\log x) = \frac{1}{x} \), we obtain: \[ \frac{dy}{dx} = \frac{\sec^2(\log x)}{x}. \] Step 2: Differentiate Again to Find \( \frac{d^2y}{dx^2} \)
Differentiating \( \frac{dy}{dx} = \frac{\sec^2(\log x)}{x} \) using the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{x \cdot \frac{d}{dx} [\sec^2(\log x)] - \sec^2(\log x) \cdot \frac{d}{dx} [x]}{x^2}. \] Using the chain rule: \[ \frac{d}{dx} [\sec^2(\log x)] = 2\sec^2(\log x) \tan(\log x) \cdot \frac{1}{x}. \] Thus, we get: \[ \frac{d^2y}{dx^2} = \frac{x \cdot \left(2\sec^2(\log x) \tan(\log x) \cdot \frac{1}{x} \right) - \sec^2(\log x)}{x^2}. \] Simplifying: \[ \frac{d^2y}{dx^2} = \frac{2\sec^2(\log x) \tan(\log x) - \sec^2(\log x)}{x^2}. \] Factoring out \( \sec^2(\log x) \): \[ \frac{d^2y}{dx^2} = \frac{\sec^2(\log x) [2\tan(\log x) - 1]}{x^2}. \] Final Answer: \( \boxed{\frac{\sec^2(\log x) [2\tan(\log x) - 1]}{x^2}} \).

Answer 2

Step 1: First derivative 

Given: \[ y = \tan(\log x) \] Use the chain rule: \[ \frac{dy}{dx} = \sec^2(\log x) \cdot \frac{d}{dx}(\log x) = \sec^2(\log x) \cdot \frac{1}{x} \] So: \[ \frac{dy}{dx} = \frac{\sec^2(\log x)}{x} \]

Step 2: Second derivative

Use the product and chain rules: \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{\sec^2(\log x)}{x} \right) \] Let’s differentiate: 
Let \( u = \sec^2(\log x) \), \( v = x \), then: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \] Now: \[ \frac{du}{dx} = \frac{d}{dx} \sec^2(\log x) = 2 \sec(\log x) \cdot \sec(\log x) \tan(\log x) \cdot \frac{1}{x} = \frac{2 \sec^2(\log x) \tan(\log x)}{x} \] So: \[ \frac{d^2y}{dx^2} = \frac{x \cdot \frac{2 \sec^2(\log x) \tan(\log x)}{x} - \sec^2(\log x)}{x^2} \] Simplify: \[ \frac{d^2y}{dx^2} = \frac{2 \sec^2(\log x) \tan(\log x) - \sec^2(\log x)}{x^2} = \frac{\sec^2(\log x)[2 \tan(\log x) - 1]}{x^2} \]

Answer:

\( \boxed{ \frac{ \sec^2(\log x)[2 \tan(\log x) - 1] }{x^2} } \)