If `x^(y)+y^(x)=1,"find "dy/dx`

Answer

To find \( \frac{dy}{dx} \) given the equation \( x^y + y^x = 1 \), we will use implicit differentiation. Let's go through the steps systematically.

### Step 1: Differentiate the equation implicitly
Given:
\[ x^y + y^x = 1 \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = \frac{d}{dx}(1) \]
This simplifies to:
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = 0 \]

### Step 2: Differentiate \( x^y \)
Using the product rule and chain rule:
\[ \frac{d}{dx}(x^y) = y \cdot x^{y-1} \cdot \frac{dx}{dx} + x^y \cdot \frac{dy}{dx} \cdot \log(x) \]
This simplifies to:
\[ y x^{y-1} + x^y \log(x) \frac{dy}{dx} \]

### Step 3: Differentiate \( y^x \)
Again, using the product rule:
\[ \frac{d}{dx}(y^x) = x \cdot y^{x-1} \cdot \frac{dy}{dx} + y^x \cdot \log(y) \]

### Step 4: Combine the derivatives
Substituting back into the differentiated equation:
\[ y x^{y-1} + x^y \log(x) \frac{dy}{dx} + \left( x y^{x-1} \frac{dy}{dx} + y^x \log(y) \right) = 0 \]

### Step 5: Rearranging the equation
Grouping the terms involving \( \frac{dy}{dx} \):
\[ y x^{y-1} + y^x \log(y) + \left( x^y \log(x) + x y^{x-1} \right) \frac{dy}{dx} = 0 \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Isolate \( \frac{dy}{dx} \):
\[ \left( x^y \log(x) + x y^{x-1} \right) \frac{dy}{dx} = -\left( y x^{y-1} + y^x \log(y) \right) \]
Thus,
\[ \frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log(y)}{x^y \log(x) + x y^{x-1}} \]

### Final Answer
\[ \frac{dy}{dx} = -\frac{y x^{y-1} + y^x \log(y)}{x^y \log(x) + x y^{x-1}} \]