Find `(dy)/(dx)` of `x^y+y^x=a^b`

Answer

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

### Step 1: Differentiate both sides with respect to \(x\)

We start with the equation:
\[ x^y + y^x = a^b \]

Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) = \frac{d}{dx}(a^b) \]

Since \(a^b\) 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:
\[ \frac{d}{dx}(x^y) = y \cdot x^{y-1} + x^y \cdot \frac{dy}{dx} \cdot \ln(x) \]

### Step 3: Differentiate \(y^x\)

Using the product rule and chain rule:
\[ \frac{d}{dx}(y^x) = y^x \cdot \ln(y) + x^{y} \cdot \frac{dy}{dx} \]

### Step 4: Combine the derivatives

Putting it all together, we have:
\[ y \cdot x^{y-1} + y^x \cdot \ln(y) + x^y \cdot \frac{dy}{dx} \cdot \ln(x) = 0 \]

### Step 5: Solve for \(\frac{dy}{dx}\)

Rearranging to isolate \(\frac{dy}{dx}\):
\[ x^y \cdot \frac{dy}{dx} \cdot \ln(x) = - (y \cdot x^{y-1} + y^x \cdot \ln(y)) \]

Now divide both sides by \(x^y \cdot \ln(x)\):
\[ \frac{dy}{dx} = -\frac{y \cdot x^{y-1} + y^x \cdot \ln(y)}{x^y \cdot \ln(x)} \]

### Final Result

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

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