Differentiate the following function with respect of `x`: `x^2sinx \ logx`

Explanation

To differentiate the function \( f(x) = x^2 \sin x \log x \) with respect to \( x \), we will use the generalized product rule. The generalized product rule states that if we have a product of multiple functions, we differentiate each function one at a time while keeping the others constant and then sum all those results.

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = x^2 \), \( v = \sin x \), and \( w = \log x \). We need to differentiate the product \( f(x) = u \cdot v \cdot w \).

2. **Apply the Generalized Product Rule**:
According to the generalized product rule:
\[ \frac{d}{dx}(u \cdot v \cdot w) = u'vw + uv'w + uvw' \]

3. **Differentiate Each Function**:
- \( u' = \frac{d}{dx}(x^2) = 2x \)
- \( v' = \frac{d}{dx}(\sin x) = \cos x \)
- \( w' = \frac{d}{dx}(\log x) = \frac{1}{x} \)

4. **Substitute the Derivatives into the Generalized Product Rule**:
Now substitute \( u, v, w, u', v', \) and \( w' \) into the formula:
\[ f'(x) = (2x)(\sin x)(\log x) + (x^2)(\cos x)(\log x) + (x^2)(\sin x)\left(\frac{1}{x}\right) \]

5. **Simplify Each Term**:
- The first term is \( 2x \sin x \log x \).
- The second term is \( x^2 \cos x \log x \).
- The third term simplifies to \( x \sin x \).

6. **Combine All Terms**:
Now combine all the terms:
\[ f'(x) = 2x \sin x \log x + x^2 \cos x \log x + x \sin x \]

7. **Factor Out Common Terms**:
We can factor out \( x \) from the expression:
\[ f'(x) = x \left( 2 \sin x \log x + x \cos x \log x + \sin x \right) \]

8. **Final Result**:
Thus, the derivative of the function \( f(x) = x^2 \sin x \log x \) is:
\[ f'(x) = x \left( 2 \sin x \log x + x \cos x \log x + \sin x \right) \]