Find `(dy)/(dx) for y=x^xdot`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = x^x\), we will follow these steps:

### Step 1: Take the logarithm of both sides
Start by taking the natural logarithm of both sides of the equation:
\[ \log y = \log(x^x) \]

### Step 2: Apply the logarithmic identity
Using the property of logarithms that states \(\log(a^b) = b \cdot \log(a)\), we can rewrite the right-hand side:
\[ \log y = x \cdot \log x \]

### Step 3: Differentiate both sides
Now, differentiate both sides with respect to \(x\). Remember to use implicit differentiation on the left side:
\[ \frac{d}{dx}(\log y) = \frac{d}{dx}(x \cdot \log x) \]

Using the chain rule on the left side:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(x \cdot \log x) \]

### Step 4: Apply the product rule on the right side
Now, apply the product rule to differentiate \(x \cdot \log x\):
\[ \frac{d}{dx}(x \cdot \log x) = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) \]

### Step 5: Compute the derivatives
The derivative of \(\log x\) is \(\frac{1}{x}\) and the derivative of \(x\) is \(1\):
\[ \frac{d}{dx}(x \cdot \log x) = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x \]

### Step 6: Substitute back into the equation
Now we substitute this back into our differentiated equation:
\[ \frac{1}{y} \cdot \frac{dy}{dx} = 1 + \log x \]

### Step 7: Solve for \(\frac{dy}{dx}\)
Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = y \cdot (1 + \log x) \]

### Step 8: Substitute \(y\) back in
Since we know that \(y = x^x\), we substitute this back into our equation:
\[ \frac{dy}{dx} = x^x \cdot (1 + \log x) \]

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

---