If `x+y=3e^2` then `D(x^y)` vanishes when x equals to

Answer

To solve the problem, we need to find the value of \( x \) for which the derivative \( D(x^y) \) vanishes, given the equation \( x + y = 3e^2 \).

### Step-by-Step Solution:

1. **Given Equation**: Start with the equation provided:
\[ x + y = 3e^2 \]

2. **Express \( y \)**: From the equation, express \( y \) in terms of \( x \):
\[ y = 3e^2 - x \]

3. **Define \( t \)**: Let \( t = x^y \). We will differentiate \( t \) with respect to \( x \).

4. **Take Logarithm**: Take the natural logarithm of both sides:
\[ \log t = y \log x \]

5. **Differentiate Both Sides**: Differentiate both sides with respect to \( x \):
\[ \frac{1}{t} \frac{dt}{dx} = \frac{dy}{dx} \log x + \frac{y}{x} \]

6. **Differentiate the Given Equation**: Differentiate \( x + y = 3e^2 \):
\[ 1 + \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -1 \]

7. **Substitute \( \frac{dy}{dx} \)**: Substitute \( \frac{dy}{dx} \) back into the differentiated equation:
\[ \frac{1}{t} \frac{dt}{dx} = -\log x + \frac{y}{x} \]

8. **Set Derivative to Zero**: Since we want \( D(x^y) \) to vanish, set \( \frac{dt}{dx} = 0 \):
\[ 0 = -\log x + \frac{y}{x} \]

9. **Rearrange the Equation**: Rearranging gives:
\[ \log x = \frac{y}{x} \]

10. **Substitute for \( y \)**: Substitute \( y = 3e^2 - x \) into the equation:
\[ \log x = \frac{3e^2 - x}{x} \]

11. **Multiply Through by \( x \)**: Multiply both sides by \( x \):
\[ x \log x = 3e^2 - x \]

12. **Rearrange**: Rearranging gives:
\[ x \log x + x = 3e^2 \]
or
\[ x(1 + \log x) = 3e^2 \]

13. **Test for \( x = e^2 \)**: Substitute \( x = e^2 \) into the equation:
\[ e^2(1 + \log e^2) = e^2(1 + 2) = 3e^2 \]

14. **Conclusion**: Since the equation holds true, we conclude that:
\[ x = e^2 \]

### Final Answer:
The value of \( x \) for which \( D(x^y) \) vanishes is:
\[ \boxed{e^2} \]