Question

If \( e^y = x^x \), then which of the following is true?

• A. \( y \frac{d^2y}{dx^2} = 1 \)
• B. \( \frac{d^2y}{dx^2} - y = 0 \)
• C. \( \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0 \)
• D. \( y \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1 = 0 \)

Answer 1

The Correct Option is D

We start with the given equation:

\( e^y = x^x \).

Take the natural logarithm of both sides:

\( y = \ln(x^x) \).

Simplify using logarithmic properties:

\( y = x \ln(x) \).

Differentiate \( y \) with respect to \( x \):

\( \frac{dy}{dx} = \ln(x) + 1 \).

Differentiate again to find the second derivative:

\( \frac{d^2y}{dx^2} = \frac{1}{x} \).

Substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the given options. For option (4):

\( y \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1 = \left( x \ln(x) \cdot \frac{1}{x} \right) - (\ln(x) + 1) + 1 \).

Simplify:

\( \ln(x) - (\ln(x) + 1) + 1 = 0 \).

Thus, option (4) satisfies the equation.