If `x^(x)+y^(y)=1` then find `(dy)/(dx)`.

Answer

To find \(\frac{dy}{dx}\) for the equation \(x^x + y^y = 1\), we will differentiate both sides with respect to \(x\) and apply implicit differentiation. Here’s a step-by-step solution:

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

### Step 2: Differentiate \(x^x\)
To differentiate \(x^x\), we can use logarithmic differentiation:
Let \(u = x^x\). Taking the natural logarithm of both sides:
\[ \ln u = x \ln x \]
Differentiating both sides with respect to \(x\):
\[ \frac{1}{u} \frac{du}{dx} = \ln x + 1 \]
Thus,
\[ \frac{du}{dx} = u(\ln x + 1) = x^x(\ln x + 1) \]

### Step 3: Differentiate \(y^y\)
Now, let \(v = y^y\). Again using logarithmic differentiation:
\[ \ln v = y \ln y \]
Differentiating both sides with respect to \(x\):
\[ \frac{1}{v} \frac{dv}{dx} = \frac{dy}{dx} \ln y + y \frac{1}{y} \frac{dy}{dx} \]
This simplifies to:
\[ \frac{1}{v} \frac{dv}{dx} = \frac{dy}{dx} (\ln y + 1) \]
Thus,
\[ \frac{dv}{dx} = v \frac{dy}{dx} (\ln y + 1) = y^y \frac{dy}{dx} (\ln y + 1) \]

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

### Step 5: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ y^y \frac{dy}{dx} (\ln y + 1) = -x^x(\ln x + 1) \]
Now, divide both sides by \(y^y(\ln y + 1)\):
\[ \frac{dy}{dx} = \frac{-x^x(\ln x + 1)}{y^y(\ln y + 1)} \]

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

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