If `y = x^(2) + x^(log x)`, then `(dy)/(dx)` is

Options

A. `2x + 2/x log x.(x^(log x))`
B. `2x + x/2 log x`
C. `2x + 2/x log x`
D. `2x + 1/x log x`

Answer

To find the derivative of the function \( y = x^2 + x^{\log x} \), we will differentiate each term separately and then combine the results.

### Step-by-Step Solution

1. **Differentiate the first term \( x^2 \)**:
\[ \frac{d}{dx}(x^2) = 2x \]

2. **Differentiate the second term \( x^{\log x} \)**:
To differentiate \( x^{\log x} \), we will use logarithmic differentiation. First, we take the natural logarithm of both sides:
\[ \log y = \log(x^{\log x}) = \log x \cdot \log x = (\log x)^2 \]

3. **Differentiate both sides with respect to \( x \)**:
Using the chain rule on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = 2 \log x \cdot \frac{1}{x} \]
This simplifies to:
\[ \frac{dy}{dx} = 2 \log x \cdot \frac{y}{x} \]

4. **Substitute \( y \) back into the equation**:
Since \( y = x^{\log x} \), we substitute this back in:
\[ \frac{dy}{dx} = 2 \log x \cdot \frac{x^{\log x}}{x} = 2 \log x \cdot x^{\log x - 1} \]

5. **Combine the derivatives**:
Now, we combine the derivatives of both terms:
\[ \frac{dy}{dx} = 2x + 2 \log x \cdot x^{\log x - 1} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 2x + 2 \log x \cdot x^{\log x - 1} \]