If `f(x)=cosxcos2xcos4xcos8xcos16x," then "f'((pi)/(4))` is

Options

A. `sqrt(2)`
B. `(1)/(sqrt(2))`
C. 1
D. none of these

Explanation

To find \( f'\left(\frac{\pi}{4}\right) \) for the function \( f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x \), we will first simplify the function and then differentiate it.

### Step 1: Rewrite the function
We can express the product of cosines in terms of sine using the double angle identity:
\[ \cos x = \frac{\sin 2x}{2 \sin x} \]
We will apply this identity iteratively to express \( f(x) \).

### Step 2: Apply the identity
1. **First application**:
\[ f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x \]
Rewrite \( \cos x \):
\[ f(x) = \frac{\sin 2x}{2 \sin x} \cos 2x \cos 4x \cos 8x \cos 16x \]

2. **Second application**:
Rewrite \( \cos 2x \):
\[ f(x) = \frac{\sin 2x}{2 \sin x} \cdot \frac{\sin 4x}{2 \sin 2x} \cos 4x \cos 8x \cos 16x \]
Simplifying gives:
\[ f(x) = \frac{\sin 4x}{4 \sin x} \cos 4x \cos 8x \cos 16x \]

3. **Continue applying the identity**:
Repeating this process, we get:
\[ f(x) = \frac{\sin 32x}{32 \sin x} \]

### Step 3: Differentiate the function
Now we differentiate \( f(x) \):
\[ f(x) = \frac{\sin 32x}{32 \sin x} \]
Using the quotient rule:
\[ f'(x) = \frac{(32 \cos 32x \cdot \sin x) - (\sin 32x \cdot 32 \cos x)}{(32 \sin x)^2} \]

### Step 4: Evaluate at \( x = \frac{\pi}{4} \)
Now we substitute \( x = \frac{\pi}{4} \):
1. Calculate \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).
2. Calculate \( 32 \cdot \frac{\pi}{4} = 8\pi \).

Now substituting into the derivative:
\[ f'\left(\frac{\pi}{4}\right) = \frac{(32 \cos(8\pi) \cdot \frac{1}{\sqrt{2}}) - (\sin(8\pi) \cdot 32 \cdot \frac{1}{\sqrt{2}})}{(32 \cdot \frac{1}{\sqrt{2}})^2} \]
Since \( \cos(8\pi) = 1 \) and \( \sin(8\pi) = 0 \):
\[ f'\left(\frac{\pi}{4}\right) = \frac{(32 \cdot 1 \cdot \frac{1}{\sqrt{2}}) - (0)}{(32 \cdot \frac{1}{\sqrt{2}})^2} \]
This simplifies to:
\[ f'\left(\frac{\pi}{4}\right) = \frac{32/\sqrt{2}}{1024/2} = \frac{32/\sqrt{2}}{512} = \frac{32}{512\sqrt{2}} = \frac{1}{16\sqrt{2}} \]

### Final Result
Thus, the value of \( f'\left(\frac{\pi}{4}\right) \) is:
\[ \frac{1}{16\sqrt{2}} \]