If `x^(y)+y^(x)=4,"find "dy/dx`

Answer

To find \(\frac{dy}{dx}\) for the equation \(x^y + y^x = 4\), we will use implicit differentiation. Here’s a step-by-step solution:

### Step 1: Differentiate both sides of the equation
We start with the equation:
\[ x^y + y^x = 4 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = \frac{d}{dx}(4) \]
Since the derivative of a constant is zero, we have:
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = 0 \]

### Step 2: Differentiate \(x^y\)
Using the product rule and chain rule:
\[ \frac{d}{dx}(x^y) = y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) \]
Here, \(y\) is treated as a function of \(x\).

### Step 3: Differentiate \(y^x\)
Again, using the product rule:
\[ \frac{d}{dx}(y^x) = y^x \cdot \ln(y) + x^{y-1} \cdot \frac{dy}{dx} \]
Here, \(y\) is treated as a function of \(x\).

### Step 4: Substitute derivatives back into the equation
Now substituting the derivatives back into the equation:
\[ y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) + y^x \cdot \ln(y) + x^{y-1} \cdot \frac{dy}{dx} = 0 \]

### Step 5: Collect all terms involving \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ x^y \cdot \frac{dy}{dx} \cdot \ln(x) + x^{y-1} \cdot \frac{dy}{dx} = -y \cdot x^{y-1} - y^x \cdot \ln(y) \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left( x^y \cdot \ln(x) + x^{y-1} \right) = -y \cdot x^{y-1} - y^x \cdot \ln(y) \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Finally, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-y \cdot x^{y-1} - y^x \cdot \ln(y)}{x^y \cdot \ln(x) + x^{y-1}} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is given by:
\[ \frac{dy}{dx} = \frac{-y \cdot x^{y-1} - y^x \cdot \ln(y)}{x^y \cdot \ln(x) + x^{y-1}} \]