If `f(x)=|[x^n, n!, 2; cosx, cos((npi)/2), 4; sinx ,sin((npi)/2), 8]|` then find the value of `(d^n)/(dx^n)([f(x)])_(x=0)dot(n in z)dot`

Explanation

To solve the problem, we need to evaluate the nth derivative of the function \( f(x) \) at \( x = 0 \), where

\[ f(x) = \left| \begin{array}{ccc} x^n & n! & 2 \\ \cos x & \cos\left(\frac{n\pi}{2}\right) & 4 \\ \sin x & \sin\left(\frac{n\pi}{2}\right) & 8 \end{array} \right| \]

### Step 1: Evaluate the Determinant

The determinant can be calculated using the formula for a 3x3 determinant:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant:

\[ f(x) = x^n \left( \cos\left(\frac{n\pi}{2}\right) \cdot 8 - 4 \cdot \sin\left(\frac{n\pi}{2}\right) \right) - n! \left( \cos x \cdot 8 - 4 \cdot \sin x \right) + 2 \left( \cos x \cdot \sin\left(\frac{n\pi}{2}\right) - \sin x \cdot \cos\left(\frac{n\pi}{2}\right) \right) \]

### Step 2: Differentiate \( f(x) \)

Now, we will differentiate \( f(x) \) with respect to \( x \) using the properties of determinants. The differentiation of a determinant can be done by differentiating each row or column, and we will apply this property iteratively.

1. The first derivative \( f'(x) \) will involve differentiating \( x^n \), \( \cos x \), and \( \sin x \).
2. The second derivative \( f''(x) \) will involve differentiating the results of \( f'(x) \), and so on until we reach the nth derivative.

### Step 3: Find \( f^{(n)}(x) \)

After differentiating \( n \) times, we will have:

\[ f^{(n)}(x) = n! \cdot \left( \text{some function of } x \right) \]

### Step 4: Evaluate at \( x = 0 \)

Now we need to evaluate \( f^{(n)}(0) \):

\[ f^{(n)}(0) = n! \cdot \left( \text{value of the function at } x = 0 \right) \]

Substituting \( x = 0 \) into the determinant gives us:

\[ f(0) = \left| \begin{array}{ccc} 0 & n! & 2 \\ 1 & \cos\left(\frac{n\pi}{2}\right) & 4 \\ 0 & \sin\left(\frac{n\pi}{2}\right) & 8 \end{array} \right| \]

### Step 5: Simplify the Determinant

The determinant simplifies to:

\[ = 0 \cdot \left( \cos\left(\frac{n\pi}{2}\right) \cdot 8 - 4 \cdot \sin\left(\frac{n\pi}{2}\right) \right) - n! \cdot (1 \cdot 8 - 4 \cdot 0) + 2 \cdot (1 \cdot \sin\left(\frac{n\pi}{2}\right) - 0 \cdot \cos\left(\frac{n\pi}{2}\right)) \]

This results in:

\[ = -n! \cdot 8 + 2 \cdot \sin\left(\frac{n\pi}{2}\right) \]

### Step 6: Final Result

Since two rows (or two columns) are similar in the determinant, the determinant evaluates to zero:

\[ f^{(n)}(0) = 0 \]

Thus, the value of \( \frac{d^n}{dx^n} f(x) \bigg|_{x=0} = 0 \).