Differentiate the following w.r.t.x.
`x^(x)`

Answer

To differentiate the function \( y = x^x \) with respect to \( x \), we can follow these steps:

### Step-by-Step Solution:

1. **Define the function**:
Let \( y = x^x \).

2. **Take the natural logarithm of both sides**:
\[ \ln(y) = \ln(x^x) \]

3. **Apply the logarithmic identity**:
Using the property of logarithms, \( \ln(a^b) = b \ln(a) \), we can rewrite the equation:
\[ \ln(y) = x \ln(x) \]

4. **Differentiate both sides with respect to \( x \)**:
We will use implicit differentiation on the left side and the product rule on the right side.
\[ \frac{d}{dx}(\ln(y)) = \frac{d}{dx}(x \ln(x)) \]

5. **Differentiate the left side**:
Using the chain rule, we have:
\[ \frac{1}{y} \frac{dy}{dx} \]

6. **Differentiate the right side using the product rule**:
Let \( u = x \) and \( v = \ln(x) \). Then,
\[ \frac{d}{dx}(x \ln(x)) = u \frac{dv}{dx} + v \frac{du}{dx} \]
where \( \frac{dv}{dx} = \frac{1}{x} \) and \( \frac{du}{dx} = 1 \).
Thus,
\[ \frac{d}{dx}(x \ln(x)) = x \cdot \frac{1}{x} + \ln(x) \cdot 1 = 1 + \ln(x) \]

7. **Set the derivatives equal**:
Now we have:
\[ \frac{1}{y} \frac{dy}{dx} = 1 + \ln(x) \]

8. **Solve for \( \frac{dy}{dx} \)**:
Multiply both sides by \( y \):
\[ \frac{dy}{dx} = y(1 + \ln(x)) \]

9. **Substitute back for \( y \)**:
Recall that \( y = x^x \):
\[ \frac{dy}{dx} = x^x(1 + \ln(x)) \]

### Final Answer:
\[ \frac{dy}{dx} = x^x(1 + \ln(x)) \]

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