When dealing with logarithmic and exponential functions, applying the natural logarithm to both sides often simplifies the expression. Remember to use the properties of logarithms, such as \( \ln(a^b) = b \ln(a) \), to simplify the equation. For differentiation, don’t forget to use the chain rule when you have composite functions, like \( x^x \). This will help you differentiate correctly and find the second derivative, which is useful for solving problems like this one.
Given the equation:
\( e^y = x^x \).
Applying the natural logarithm to both sides yields:
\( y = \ln(x^x) \).
Simplifying with logarithmic properties results in:
\( y = x \ln(x) \).
Differentiating \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \ln(x) + 1 \).
A second differentiation yields the second derivative:
\( \frac{d^2y}{dx^2} = \frac{1}{x} \).
Substituting \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the provided options. For option (4):
\( y \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1 = \left( x \ln(x) \cdot \frac{1}{x} \right) - (\ln(x) + 1) + 1 \).
Simplification leads to:
\( \ln(x) - (\ln(x) + 1) + 1 = 0 \).
Therefore, option (4) satisfies the equation.