If `f(x) = cos x\ cos 2x\ cos 2^2\ x\ cos 2^3 x\ ....cos2^(n-1) x` and `n gt 1,` then `f'(pi/2)` is

Options

A. 1
B. 0
C. -1
D. (-1)^n-1

Explanation

To solve the problem, we need to differentiate the function \( f(x) = \cos x \cos 2x \cos 2^2 x \cos 2^3 x \ldots \cos 2^{n-1} x \) and evaluate \( f'(\frac{\pi}{2}) \).

### Step 1: Rewrite the function
We can express \( f(x) \) as:
\[ f(x) = \prod_{k=0}^{n-1} \cos(2^k x) \]

### Step 2: Use the product rule
To differentiate \( f(x) \), we will use the product rule. The derivative of a product of functions is given by:
\[ (fg)' = f'g + fg' \]
For \( f(x) \), we have:
\[ f'(x) = \sum_{k=0}^{n-1} \left( \prod_{j=0, j \neq k}^{n-1} \cos(2^j x) \right) \cdot \frac{d}{dx}[\cos(2^k x)] \]

### Step 3: Differentiate \( \cos(2^k x) \)
The derivative of \( \cos(2^k x) \) is:
\[ \frac{d}{dx}[\cos(2^k x)] = -\sin(2^k x) \cdot 2^k \]
Thus, substituting this into our derivative expression, we get:
\[ f'(x) = \sum_{k=0}^{n-1} \left( \prod_{j=0, j \neq k}^{n-1} \cos(2^j x) \right) \cdot (-\sin(2^k x) \cdot 2^k) \]

### Step 4: Evaluate at \( x = \frac{\pi}{2} \)
Now we need to evaluate \( f'(\frac{\pi}{2}) \). First, we observe that:
\[ \cos(2^k \cdot \frac{\pi}{2}) = 0 \quad \text{for odd } k \]
and
\[ \cos(2^k \cdot \frac{\pi}{2}) = (-1)^{m} \quad \text{for even } k \text{ where } m = \frac{2^k}{2} \]

### Step 5: Simplifying \( f'(\frac{\pi}{2}) \)
At \( x = \frac{\pi}{2} \):
- For \( k = 0 \), \( \sin(2^0 \cdot \frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 \)
- For \( k = 1 \), \( \sin(2^1 \cdot \frac{\pi}{2}) = \sin(\pi) = 0 \)
- For \( k = 2 \), \( \sin(2^2 \cdot \frac{\pi}{2}) = \sin(2\pi) = 0 \)
- Continuing this, we find that all terms for \( k \geq 1 \) will contribute 0 due to the sine terms.

Thus, we only need to consider the term when \( k = 0 \):
\[ f'(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) \cdot 2^0 \cdot \prod_{j=1}^{n-1} \cos(2^j \cdot \frac{\pi}{2}) = -1 \cdot \prod_{j=1}^{n-1} 0 = 0 \]

### Final Result
So, we conclude that:
\[ f'(\frac{\pi}{2}) = 0 \]