If `f(x)` is given by
`f(x)=(cosx+i sinx)(cos3x+isin3x)`.........
.......`[cos(2n-1)x+isin(2n-1)x]`,
then f''(x) is equal to

Options

A. `n^(3)f(x)`
B. `-n^(4)f(x)`
C. `-n^(2)f(x)`
D. `n^(4)f(x)`

Answer

To solve the problem, we need to find the second derivative \( f''(x) \) of the function

\[ f(x) = ( \cos x + i \sin x )( \cos 3x + i \sin 3x ) \cdots ( \cos (2n-1)x + i \sin (2n-1)x ) \]

### Step 1: Rewrite the function using Euler's formula

Using Euler's formula, we can express \( \cos x + i \sin x \) as \( e^{ix} \). Therefore, we can rewrite \( f(x) \) as:

\[ f(x) = e^{ix} \cdot e^{i3x} \cdots e^{i(2n-1)x} \]

### Step 2: Combine the exponents

Since the bases are the same (all are \( e \)), we can add the exponents:

\[ f(x) = e^{i(x + 3x + 5x + \ldots + (2n-1)x)} \]

### Step 3: Find the sum of the series

The series \( x + 3x + 5x + \ldots + (2n-1)x \) is an arithmetic series where the first term \( a = x \) and the common difference \( d = 2x \). The number of terms \( n \) is \( n \).

The sum of the first \( n \) terms of an arithmetic series is given by:

\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]

Substituting the values:

\[ S_n = \frac{n}{2} \cdot (2x + (n-1) \cdot 2x) = \frac{n}{2} \cdot (2nx) = n^2 x \]

### Step 4: Substitute the sum back into the function

Thus, we can rewrite \( f(x) \) as:

\[ f(x) = e^{in^2 x} \]

### Step 5: Differentiate to find \( f'(x) \)

Now we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} e^{in^2 x} = in^2 e^{in^2 x} \]

### Step 6: Differentiate again to find \( f''(x) \)

Now we differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx} (in^2 e^{in^2 x}) = in^2 \cdot in^2 e^{in^2 x} = -n^4 e^{in^2 x} \]

### Step 7: Final expression for \( f''(x) \)

Thus, we have:

\[ f''(x) = -n^4 e^{in^2 x} \]

### Conclusion

The final result for \( f''(x) \) is:

\[ f''(x) = -n^4 f(x) \]