Cosider the function `f(x)=|{:(x^(3),sin x,cos x),(6,-1,0),(p,p^(2),p^(3)):}|`, where p is a constant.
What is the value of f'(0) ?

Options

A. `p^(3)`
B. `3p^(3)`
C. `6p^(3)`
D. `-6p^(3)`

Explanation

To solve the problem, we need to find the derivative of the function \( f(x) \) defined as the determinant of a matrix, and then evaluate it at \( x = 0 \).

### Step-by-step Solution:

1. **Define the function**:
The function is given as:
\[ f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \]

2. **Differentiate \( f(x) \)**:
We need to differentiate the determinant with respect to \( x \). We will use the property of determinants that allows us to differentiate each row independently.

- **Differentiate the first row**:
\[ \frac{d}{dx}(x^3) = 3x^2, \quad \frac{d}{dx}(\sin x) = \cos x, \quad \frac{d}{dx}(\cos x) = -\sin x \]
Thus, the derivative of the first row is:
\[ \begin{vmatrix} 3x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \]

- **Differentiate the second row**:
Since the second row contains constants, its derivative is zero:
\[ \begin{vmatrix} x^3 & \sin x & \cos x \\ 0 & 0 & 0 \\ p & p^2 & p^3 \end{vmatrix} \]

- **Differentiate the third row**:
Similarly, the third row also contains constants, so its derivative is zero:
\[ \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ 0 & 0 & 0 \end{vmatrix} \]

Therefore, we can express \( f'(x) \) as:
\[ f'(x) = \begin{vmatrix} 3x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \]

3. **Evaluate \( f'(0) \)**:
Now we substitute \( x = 0 \) into \( f'(x) \):
\[ f'(0) = \begin{vmatrix} 3(0)^2 & \cos(0) & -\sin(0) \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} = \begin{vmatrix} 0 & 1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix} \]

4. **Calculate the determinant**:
We can expand the determinant along the first row:
\[ f'(0) = 0 \cdot \begin{vmatrix} -1 & 0 \\ p^2 & p^3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} + 0 \cdot \begin{vmatrix} 6 & -1 \\ p & p^2 \end{vmatrix} \]
The first and third terms are zero, so we only need to calculate the second term:
\[ = -\begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} = - (6 \cdot p^3 - 0) = -6p^3 \]

5. **Final Result**:
Thus, the value of \( f'(0) \) is:
\[ f'(0) = -6p^3 \]