if f(x) = cos x cos 2x cos 4x cos 8 x cos 16 x , then ` f' (pi //4)` is

Options

A. `1`
B. ` sqrt(2)`
C. `1 // sqrt(2)`
D. None

Answer

To find \( f'(\frac{\pi}{4}) \) for the function

\[ f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x, \]

we will use the product rule and some trigonometric identities to simplify our calculations.

### Step 1: Rewrite the function
We can rewrite the function using the identity \( \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \). However, for simplicity, we will use a different approach by multiplying and dividing by \( 2\sin x \).

\[ f(x) = \cos x \cos 2x \cos 4x \cos 8x \cos 16x \cdot \frac{2\sin x}{2\sin x} = \frac{2\sin x \cos x \cos 2x \cos 4x \cos 8x \cos 16x}{2\sin x}. \]

### Step 2: Apply the double angle formula
Using the double angle formula, we can simplify the numerator step by step:

1. \( 2\sin x \cos x = \sin 2x \)
2. \( 2\sin 2x \cos 2x = \sin 4x \)
3. \( 2\sin 4x \cos 4x = \sin 8x \)
4. \( 2\sin 8x \cos 8x = \sin 16x \)

Continuing this process, we find that:

\[ f(x) = \frac{\sin 32x}{2^5 \sin x}. \]

### Step 3: Differentiate \( f(x) \)
Now we differentiate \( f(x) \) using the quotient rule:

\[ f'(x) = \frac{(2^5 \sin x)(\frac{d}{dx}(\sin 32x)) - (\sin 32x)(\frac{d}{dx}(2^5 \sin x))}{(2^5 \sin x)^2}. \]

Calculating the derivatives:

- The derivative of \( \sin 32x \) is \( 32 \cos 32x \).
- The derivative of \( \sin x \) is \( \cos x \).

Substituting these into our expression gives:

\[ f'(x) = \frac{(2^5 \sin x)(32 \cos 32x) - (\sin 32x)(2^5 \cos x)}{(2^5 \sin x)^2}. \]

### Step 4: Evaluate at \( x = \frac{\pi}{4} \)
Now we substitute \( x = \frac{\pi}{4} \):

1. Calculate \( \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).
2. Calculate \( \sin 32 \cdot \frac{\pi}{4} = \sin 8\pi = 0 \).
3. Calculate \( \cos 32 \cdot \frac{\pi}{4} = \cos 8\pi = 1 \).

Substituting these values into our derivative:

\[ f'(\frac{\pi}{4}) = \frac{(2^5 \cdot \frac{1}{\sqrt{2}})(32 \cdot 1) - (0)(2^5 \cdot \cos \frac{\pi}{4})}{(2^5 \cdot \frac{1}{\sqrt{2}})^2}. \]

This simplifies to:

\[ f'(\frac{\pi}{4}) = \frac{(32 \cdot 32)}{(2^5 \cdot \frac{1}{\sqrt{2}})^2} = \frac{1024}{32} = 32. \]

### Final Result
Thus, the value of \( f'(\frac{\pi}{4}) \) is

\[ \boxed{32}. \]