If `x^(y)+y^(x)+x^(x)=m^(n)." then find the value of "(dy)/(dx)`.

Answer

To find the value of \(\frac{dy}{dx}\) given the equation \(x^y + y^x + x^x = m^n\), we will differentiate both sides of the equation with respect to \(x\). Let's go through the solution step by step.

### Step 1: Differentiate both sides
We start with the equation:
\[ x^y + y^x + x^x = m^n \]
Since \(m\) and \(n\) are constants, the derivative of \(m^n\) with respect to \(x\) is \(0\). Therefore, we differentiate the left side:
\[ \frac{d}{dx}(x^y) + \frac{d}{dx}(y^x) + \frac{d}{dx}(x^x) = 0 \]

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

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

### Step 4: Differentiate \(x^x\)
Using the chain rule:
\[ \frac{d}{dx}(x^x) = x^x \cdot (\ln(x) + 1) \]

### Step 5: Combine the derivatives
Now, substituting back into the equation:
\[ y \cdot x^{y-1} + y^x \cdot \ln(y) \cdot \frac{dy}{dx} + x^x \cdot (\ln(x) + 1) = 0 \]

### Step 6: Rearrange to isolate \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ y^x \cdot \ln(y) \cdot \frac{dy}{dx} = -y \cdot x^{y-1} - x^x \cdot (\ln(x) + 1) \]
\[ \frac{dy}{dx} = \frac{-y \cdot x^{y-1} - x^x \cdot (\ln(x) + 1)}{y^x \cdot \ln(y)} \]

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