If `f(x)=(cosx+isinx)(cos2x+isin2x)(cos3x+isin3x)......(cosn x+isinn x)` and `f(1)=1`,then `f^('')(1)` is equal to

Options

A. (a) `(n(n+1))/2`
B. (b) `{(n(n+1))/2}^2`
C. (C) `-{(n(n+1))/2}^2`
D. (d) none of these

Explanation

To solve the problem, we start with the function given:

\[ f(x) = (cos x + i sin x)(cos 2x + i sin 2x)(cos 3x + i sin 3x) \cdots (cos nx + i sin nx) \]

### Step 1: Rewrite using Euler's Formula
Using Euler's formula, we can rewrite each term \( cos kx + i sin kx \) as \( e^{ikx} \). Therefore, we have:

\[ f(x) = e^{ix} \cdot e^{i2x} \cdot e^{i3x} \cdots e^{inx} \]
...