`"Let "f(x)=|{:(cos(x+x^(2)),sin (x+x^(2)),-cos(x+x^(2))),(sin (x-x^(2)),cos (x-x^(2)),sin (x-x^(2))),(sin 2x, 0, sin (2x^(2))):}|.`
Find the value of f'(0).

Explanation

To find the value of \( f'(0) \) for the given function

\[ f(x) = \begin{vmatrix} \cos(x + x^2) & \sin(x + x^2) & -\cos(x + x^2) \\ \sin(x - x^2) & \cos(x - x^2) & \sin(x - x^2) \\ \sin(2x) & 0 & \sin(2x^2) \end{vmatrix}, \]

we will follow these steps:

### Step 1: Compute \( f(0) \)

First, we need to evaluate \( f(0) \):

\[ f(0) = \begin{vmatrix} \cos(0 + 0^2) & \sin(0 + 0^2) & -\cos(0 + 0^2) \\ \sin(0 - 0^2) & \cos(0 - 0^2) & \sin(0 - 0^2) \\ \sin(2 \cdot 0) & 0 & \sin(2 \cdot 0^2) \end{vmatrix} \]

This simplifies to:

\[ f(0) = \begin{vmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{vmatrix} \]

The determinant of a matrix with a row of zeros is zero:

\[ f(0) = 0 \]

### Step 2: Compute \( f'(x) \)

To find \( f'(x) \), we will use the property of determinants that allows us to differentiate under the determinant sign. We differentiate each row of the determinant.

Using the formula for the derivative of a determinant, we have:

\[ f'(x) = \begin{vmatrix} \frac{d}{dx}(\cos(x + x^2)) & \frac{d}{dx}(\sin(x + x^2)) & \frac{d}{dx}(-\cos(x + x^2)) \\ \frac{d}{dx}(\sin(x - x^2)) & \frac{d}{dx}(\cos(x - x^2)) & \frac{d}{dx}(\sin(x - x^2)) \\ \frac{d}{dx}(\sin(2x)) & 0 & \frac{d}{dx}(\sin(2x^2)) \end{vmatrix} \]

### Step 3: Differentiate each term

Now we compute the derivatives:

1. For the first row:
- \( \frac{d}{dx}(\cos(x + x^2)) = -\sin(x + x^2)(1 + 2x) \)
- \( \frac{d}{dx}(\sin(x + x^2)) = \cos(x + x^2)(1 + 2x) \)
- \( \frac{d}{dx}(-\cos(x + x^2)) = \sin(x + x^2)(1 + 2x) \)

2. For the second row:
- \( \frac{d}{dx}(\sin(x - x^2)) = \cos(x - x^2)(1 - 2x) \)
- \( \frac{d}{dx}(\cos(x - x^2)) = -\sin(x - x^2)(1 - 2x) \)
- \( \frac{d}{dx}(\sin(x - x^2)) = \cos(x - x^2)(1 - 2x) \)

3. For the third row:
- \( \frac{d}{dx}(\sin(2x)) = 2\cos(2x) \)
- \( \frac{d}{dx}(0) = 0 \)
- \( \frac{d}{dx}(\sin(2x^2)) = 2x\cos(2x^2) \)

### Step 4: Substitute \( x = 0 \)

Now substituting \( x = 0 \) into the derivatives:

\[ f'(0) = \begin{vmatrix} -\sin(0)(1 + 0) & \cos(0)(1 + 0) & \sin(0)(1 + 0) \\ \cos(0)(1 - 0) & -\sin(0)(1 - 0) & \cos(0)(1 - 0) \\ 2\cos(0) & 0 & 0 \end{vmatrix} \]

This simplifies to:

\[ f'(0) = \begin{vmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 2 & 0 & 0 \end{vmatrix} \]

### Step 5: Calculate the determinant

Calculating this determinant:

\[ f'(0) = 0 \cdot \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} + 0 \cdot \begin{vmatrix} 1 & 0 \\ 2 & 0 \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ \begin{vmatrix} 1 & 1 \\ 2 & 0 \end{vmatrix} = (1)(0) - (1)(2) = -2 \]

Thus,

\[ f'(0) = -(-2) = 2 \]

### Final Answer

Therefore, the value of \( f'(0) \) is:

\[ \boxed{2} \]