Consider the function
`f(x)=|{:(x^(3),sinx,cosx),(6,-1,0),(p,p,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)`

Answer

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

\[ f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p & p^3 \end{vmatrix} \]

we will follow these steps:

### Step 1: Differentiate the Determinant
To differentiate the determinant with respect to \( x \), we apply the rule for differentiating determinants. The derivative of a determinant can be found by differentiating each row while keeping the other rows constant.

\[ f'(x) = \begin{vmatrix} \frac{d}{dx}(x^3) & \frac{d}{dx}(\sin x) & \frac{d}{dx}(\cos x) \\ 6 & -1 & 0 \\ p & p & p^3 \end{vmatrix} + \begin{vmatrix} x^3 & \sin x & \cos x \\ \frac{d}{dx}(6) & \frac{d}{dx}(-1) & \frac{d}{dx}(0) \\ p & p & p^3 \end{vmatrix} + \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ \frac{d}{dx}(p) & \frac{d}{dx}(p) & \frac{d}{dx}(p^3) \end{vmatrix} \]

### Step 2: Calculate the Derivatives
Calculating the derivatives of the first row:
- \( \frac{d}{dx}(x^3) = 3x^2 \)
- \( \frac{d}{dx}(\sin x) = \cos x \)
- \( \frac{d}{dx}(\cos x) = -\sin x \)

The second row is constant, so its derivative is zero. The third row contains \( p \), which is a constant, thus its derivative is also zero.

So, we have:

\[ f'(x) = \begin{vmatrix} 3x^2 & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p & p^3 \end{vmatrix} \]

### Step 3: Evaluate \( f'(0) \)
Now we substitute \( x = 0 \):

\[ f'(0) = \begin{vmatrix} 3(0)^2 & \cos(0) & -\sin(0) \\ 6 & -1 & 0 \\ p & p & p^3 \end{vmatrix} = \begin{vmatrix} 0 & 1 & 0 \\ 6 & -1 & 0 \\ p & p & p^3 \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we compute the determinant:

\[ f'(0) = 0 \cdot \begin{vmatrix} -1 & 0 \\ p & p^3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} + 0 \cdot \begin{vmatrix} 6 & -1 \\ p & p \end{vmatrix} \]

Calculating the second determinant:

\[ \begin{vmatrix} 6 & 0 \\ p & p^3 \end{vmatrix} = 6p^3 - 0 = 6p^3 \]

Thus,

\[ f'(0) = -6p^3 \]

### Final Answer
The value of \( f'(0) \) is

\[ \boxed{-6p^3} \]