Question

If \[ y = \left( \frac{x^2 + 1}{x} \right)^x \text{ and } \frac{dy}{dx} = y \left[ g(x) + \log \left( \frac{x^2}{x+1} \right) \right], \text{ then } g(x) = \]

• A. \( \frac{x + 2}{x + 1} \)
• B. \( x \log \left( \frac{x^2}{x + 1} \right) \)
• C. \( \frac{x^2}{x + 1} \)
• D. \( \frac{x - 1}{x + 2} \)

Hint:

When using logarithmic differentiation, differentiate both sides after taking the logarithm of the expression to simplify the calculation.

Answer 1

The Correct Option is A

Step 1: Use the given information.
We are given \( y = \left( \frac{x^2 + 1}{x} \right)^x \) and the equation for \( \frac{dy}{dx} \). We can start by differentiating \( y \) using logarithmic differentiation to find \( \frac{dy}{dx} \).
Step 2: Apply the differentiation.
Differentiating the expression for \( y \), we find the value of \( g(x) \) as: \[ g(x) = \frac{x + 2}{x + 1} \]
Step 3: Conclusion.
Thus, \( g(x) = \frac{x + 2}{x + 1} \), corresponding to option (A).