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.
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.
We start with the given equation:
\[ e^y = x^x \]
Step 1: Take the natural logarithm of both sides:
\[ y = \ln(x^x) \]Step 2: Simplify using logarithmic properties:
Using the property \( \ln(a^b) = b \ln(a) \), we simplify: \[ y = x \ln(x) \]Step 3: Differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \ln(x) + 1 \]Step 4: Differentiate again to find the second derivative:
\[ \frac{d^2y}{dx^2} = \frac{1}{x} \]Step 5: Substitute \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) into the given options. For option (4):
\[ y \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1 = \left( x \ln(x) \cdot \frac{1}{x} \right) - (\ln(x) + 1) + 1 \]Step 6: Simplify the expression:
\[ \ln(x) - (\ln(x) + 1) + 1 = 0 \]Conclusion: Thus, option (4) satisfies the equation.