If `y=cos x cos 2x cos 4x cos 8x`, then `(dy)/(dx)" at "x=(pi)/(2)` is

Options

A. 1
B. 0
C. 4
D. 16

Answer

To find \(\frac{dy}{dx}\) at \(x = \frac{\pi}{2}\) for the function \(y = \cos x \cos 2x \cos 4x \cos 8x\), we will use the product rule and the chain rule of differentiation.

### Step-by-step Solution:

1. **Identify the function:**
\[ y = \cos x \cdot \cos 2x \cdot \cos 4x \cdot \cos 8x \]

2. **Differentiate using the product rule:**
The product rule states that if \(y = u_1 u_2 u_3 u_4\), then:
\[ \frac{dy}{dx} = u_1' u_2 u_3 u_4 + u_1 u_2' u_3 u_4 + u_1 u_2 u_3' u_4 + u_1 u_2 u_3 u_4' \]
where \(u_1 = \cos x\), \(u_2 = \cos 2x\), \(u_3 = \cos 4x\), and \(u_4 = \cos 8x\).

3. **Calculate the derivatives:**
- \(u_1' = -\sin x\)
- \(u_2' = -2\sin 2x\)
- \(u_3' = -4\sin 4x\)
- \(u_4' = -8\sin 8x\)

4. **Substituting into the product rule:**
\[ \frac{dy}{dx} = (-\sin x) \cos 2x \cos 4x \cos 8x + \cos x (-2\sin 2x) \cos 4x \cos 8x + \cos x \cos 2x (-4\sin 4x) \cos 8x + \cos x \cos 2x \cos 4x (-8\sin 8x) \]

5. **Evaluate at \(x = \frac{\pi}{2}\):**
- \(\cos\left(\frac{\pi}{2}\right) = 0\)
- \(\cos\left(2 \cdot \frac{\pi}{2}\right) = \cos(\pi) = -1\)
- \(\cos\left(4 \cdot \frac{\pi}{2}\right) = \cos(2\pi) = 1\)
- \(\cos\left(8 \cdot \frac{\pi}{2}\right) = \cos(4\pi) = 1\)

Thus, substituting these values:
\[ \frac{dy}{dx} = (-\sin(\frac{\pi}{2}) \cdot (-1) \cdot 1 \cdot 1) + (0) + (0) + (0) \]
\[ = -1 \cdot 1 \cdot 1 \cdot 1 = -1 \]

6. **Final Result:**
\[ \frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = -1 \]