If ` x^y +y^x =a^6 `, then ` (dy)/(dx)`= . . .. . .

Answer

To find \(\frac{dy}{dx}\) for the equation \(x^y + y^x = a^6\), 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:
\[ x^y + y^x = a^6 \]
Differentiating both sides with respect to \(x\) gives:
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = \frac{d}{dx}(a^6) \]
Since \(a^6\) is a constant, its derivative is \(0\):
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = 0 \]

### Step 2: Differentiate \(x^y\)
Using implicit differentiation and the chain rule for \(x^y\):
\[ \frac{d}{dx}(x^y) = y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) \]
So, we have:
\[ y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) \]

### Step 3: Differentiate \(y^x\)
Now, differentiate \(y^x\):
\[ \frac{d}{dx}(y^x) = y^x \cdot \ln(y) + x^{y-1} \cdot \frac{dy}{dx} \]
Thus, we have:
\[ y^x \cdot \ln(y) + x^y \cdot \frac{dy}{dx} \]

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

### Step 5: Factor out \(\frac{dy}{dx}\)
Rearranging gives:
\[ x^y \cdot \frac{dy}{dx} \cdot \ln(x) + y^x \cdot \frac{dy}{dx} = -y \cdot x^{y-1} - y^x \cdot \ln(y) \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (x^y \cdot \ln(x) + y^x) = -y \cdot x^{y-1} - y^x \cdot \ln(y) \]

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

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