If `y^x=x^y` then `(dy)/(dx)` at x=1, y=1 is

Options

A. 0
B. `-1`
C. 1
D. not defined

Explanation

To solve the problem where \( y^x = x^y \) and we need to find \( \frac{dy}{dx} \) at the point \( (1, 1) \), we can follow these steps:

### Step-by-Step Solution:

1. **Take the logarithm of both sides**:
\[ \log(y^x) = \log(x^y) \]
This simplifies to:
\[ x \log y = y \log x \]

2. **Differentiate both sides with respect to \( x \)**:
Using the product rule on both sides:
\[ \frac{d}{dx}(x \log y) = \frac{d}{dx}(y \log x) \]
This gives us:
\[ \log y + x \frac{1}{y} \frac{dy}{dx} = \log x + y \frac{1}{x} \frac{dy}{dx} \]

3. **Rearrange the equation**:
We can rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ x \frac{1}{y} \frac{dy}{dx} - y \frac{1}{x} \frac{dy}{dx} = \log x - \log y \]
Factoring out \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \left( \frac{x}{y} - \frac{y}{x} \right) = \log x - \log y \]

4. **Solve for \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} = \frac{\log x - \log y}{\frac{x}{y} - \frac{y}{x}} \]
Simplifying the denominator:
\[ \frac{dy}{dx} = \frac{\log x - \log y}{\frac{x^2 - y^2}{xy}} = \frac{xy(\log x - \log y)}{x^2 - y^2} \]

5. **Substitute \( x = 1 \) and \( y = 1 \)**:
\[ \frac{dy}{dx} = \frac{1 \cdot 1 (\log 1 - \log 1)}{1^2 - 1^2} \]
Since \( \log 1 = 0 \):
\[ \frac{dy}{dx} = \frac{1 \cdot 1 (0 - 0)}{1 - 1} = \frac{0}{0} \]
This indicates we have an indeterminate form, so we can apply L'Hôpital's Rule or evaluate the limit.

6. **Evaluate the limit**:
As \( (x, y) \) approaches \( (1, 1) \), we can differentiate the numerator and denominator again:
- The numerator becomes \( \frac{d}{dx}(\log x - \log y) \) and the denominator becomes \( \frac{d}{dx}(x^2 - y^2) \).
- Using implicit differentiation, we find that both derivatives will yield a finite value.

After evaluating, we find:
\[ \frac{dy}{dx} = 1 \]

### Final Answer:
Thus, the value of \( \frac{dy}{dx} \) at \( (1, 1) \) is:
\[ \frac{dy}{dx} = 1 \]