Question:

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

Show Hint

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.

Updated On: Mar 28, 2025
  • \( y \frac{d^2y}{dx^2} = 1 \)
  • \( \frac{d^2y}{dx^2} - y = 0 \)
  • \( \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0 \)
  • \( y \frac{d^2y}{dx^2} - \frac{dy}{dx} + 1 = 0 \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Approach Solution - 1

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.

Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

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.

Was this answer helpful?
0
0