`y=x^(logx)+(logx)^(x)`

Answer

To find the derivative of the function \( y = x^{\log x} + (\log x)^x \), we will break it down into two parts and use the properties of logarithms and differentiation.

### Step-by-Step Solution:

1. **Identify the components of the function**:
Let \( y_1 = x^{\log x} \) and \( y_2 = (\log x)^x \). Therefore, we can express \( y \) as:
\[ y = y_1 + y_2 \]

2. **Differentiate \( y \)**:
Using the sum rule of differentiation:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]

3. **Differentiate \( y_1 = x^{\log x} \)**:
To differentiate \( y_1 \), we can use logarithmic differentiation:
\[ \log y_1 = \log(x^{\log x}) = \log x \cdot \log x = (\log x)^2 \]
Now, differentiate both sides:
\[ \frac{1}{y_1} \frac{dy_1}{dx} = 2 \log x \cdot \frac{1}{x} \]
Therefore,
\[ \frac{dy_1}{dx} = y_1 \cdot \frac{2 \log x}{x} \]
Substituting back \( y_1 = x^{\log x} \):
\[ \frac{dy_1}{dx} = x^{\log x} \cdot \frac{2 \log x}{x} = 2 x^{\log x - 1} \log x \]

4. **Differentiate \( y_2 = (\log x)^x \)**:
Again, we use logarithmic differentiation:
\[ \log y_2 = x \log(\log x) \]
Differentiate both sides:
\[ \frac{1}{y_2} \frac{dy_2}{dx} = \log(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \log(\log x) + \frac{1}{\log x} \]
Therefore,
\[ \frac{dy_2}{dx} = y_2 \left( \log(\log x) + \frac{1}{\log x} \right) \]
Substituting back \( y_2 = (\log x)^x \):
\[ \frac{dy_2}{dx} = (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) \]

5. **Combine the derivatives**:
Now, we can combine the derivatives:
\[ \frac{dy}{dx} = 2 x^{\log x - 1} \log x + (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) \]

### Final Answer:
\[ \frac{dy}{dx} = 2 x^{\log x - 1} \log x + (\log x)^x \left( \log(\log x) + \frac{1}{\log x} \right) \]