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

Answer

To solve the problem \( x^y + y^x = \log a \) and find \( \frac{dy}{dx} \), we will use implicit differentiation. Here are the steps:

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

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

### Step 3: Differentiate \( y^x \)
Using the product rule and chain rule:
\[ \frac{d}{dx}(y^x) = y^x \ln(y) + x y^{x-1} \frac{dy}{dx} \]

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

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

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, isolating \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{- (yx^{y-1} + y^x \ln(y))}{x^y \ln(x) + x y^{x-1}} \]

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