if y =` sin 2x , " then " (d^(6)y)/(dx^(6)) " at" x = pi/2` is equal to

Options

A. `-64`
B. 0
C. 64
D. 128

Answer

To solve the problem of finding \(\frac{d^6y}{dx^6}\) at \(x = \frac{\pi}{2}\) for the function \(y = \sin(2x)\), we will differentiate the function six times and then evaluate the sixth derivative at the specified point.

### Step-by-Step Solution:

1. **Given Function**:
\[ y = \sin(2x) \]

2. **First Derivative**:
\[ \frac{dy}{dx} = \cos(2x) \cdot \frac{d(2x)}{dx} = 2 \cos(2x) \]

3. **Second Derivative**:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2 \cos(2x)) = 2 \cdot (-\sin(2x)) \cdot \frac{d(2x)}{dx} = -4 \sin(2x) \]

4. **Third Derivative**:
\[ \frac{d^3y}{dx^3} = \frac{d}{dx}(-4 \sin(2x)) = -4 \cos(2x) \cdot \frac{d(2x)}{dx} = -8 \cos(2x) \]

5. **Fourth Derivative**:
\[ \frac{d^4y}{dx^4} = \frac{d}{dx}(-8 \cos(2x)) = -8 \cdot (-\sin(2x)) \cdot \frac{d(2x)}{dx} = 16 \sin(2x) \]

6. **Fifth Derivative**:
\[ \frac{d^5y}{dx^5} = \frac{d}{dx}(16 \sin(2x)) = 16 \cos(2x) \cdot \frac{d(2x)}{dx} = 32 \cos(2x) \]

7. **Sixth Derivative**:
\[ \frac{d^6y}{dx^6} = \frac{d}{dx}(32 \cos(2x)) = 32 \cdot (-\sin(2x)) \cdot \frac{d(2x)}{dx} = -64 \sin(2x) \]

8. **Evaluate at \(x = \frac{\pi}{2}\)**:
\[ \frac{d^6y}{dx^6} \bigg|_{x = \frac{\pi}{2}} = -64 \sin(2 \cdot \frac{\pi}{2}) = -64 \sin(\pi) \]
Since \(\sin(\pi) = 0\):
\[ \frac{d^6y}{dx^6} \bigg|_{x = \frac{\pi}{2}} = -64 \cdot 0 = 0 \]

### Final Answer:
\[ \frac{d^6y}{dx^6} \bigg|_{x = \frac{\pi}{2}} = 0 \]