Find the derivative of the following :
`x^(3)sinx+2xcosx-2xsinx`

Answer

To find the derivative of the function \( f(x) = x^3 \sin x + 2x \cos x - 2x \sin x \), we will differentiate each term separately using the product rule and sum the results.

### Step-by-step Solution:

1. **Identify the function**:
\[ f(x) = x^3 \sin x + 2x \cos x - 2x \sin x \]

2. **Differentiate the first term \( x^3 \sin x \)**:
- Use the product rule: If \( u = x^3 \) and \( v = \sin x \), then
\[ \frac{d}{dx}(uv) = u'v + uv' \]
- Calculate \( u' = \frac{d}{dx}(x^3) = 3x^2 \) and \( v' = \frac{d}{dx}(\sin x) = \cos x \).
- Applying the product rule:
\[ \frac{d}{dx}(x^3 \sin x) = (3x^2)(\sin x) + (x^3)(\cos x) = 3x^2 \sin x + x^3 \cos x \]

3. **Differentiate the second term \( 2x \cos x \)**:
- Again, use the product rule: Let \( u = 2x \) and \( v = \cos x \).
- Calculate \( u' = \frac{d}{dx}(2x) = 2 \) and \( v' = \frac{d}{dx}(\cos x) = -\sin x \).
- Applying the product rule:
\[ \frac{d}{dx}(2x \cos x) = (2)(\cos x) + (2x)(-\sin x) = 2 \cos x - 2x \sin x \]

4. **Differentiate the third term \( -2x \sin x \)**:
- Use the product rule: Let \( u = -2x \) and \( v = \sin x \).
- Calculate \( u' = \frac{d}{dx}(-2x) = -2 \) and \( v' = \frac{d}{dx}(\sin x) = \cos x \).
- Applying the product rule:
\[ \frac{d}{dx}(-2x \sin x) = (-2)(\sin x) + (-2x)(\cos x) = -2 \sin x - 2x \cos x \]

5. **Combine all the derivatives**:
\[ f'(x) = (3x^2 \sin x + x^3 \cos x) + (2 \cos x - 2x \sin x) + (-2 \sin x - 2x \cos x) \]

6. **Simplify the expression**:
- Combine like terms:
\[ f'(x) = x^3 \cos x + 3x^2 \sin x + 2 \cos x - 2x \sin x - 2 \sin x - 2x \cos x \]
- Rearranging gives:
\[ f'(x) = x^3 \cos x + (3x^2 - 2x - 2) \sin x + (2 - 2x) \cos x \]

### Final Answer:
\[ f'(x) = x^3 \cos x + (3x^2 - 2x - 2) \sin x + (2 - 2x) \cos x \]