If ` y = sin^(n) x cos nx, " then" (dy)/(dx)` is

Options

A. ` n sin^(n-1) x sin (n+1)x `
B. `n sin^(n-1) x cos (n-1) x `
C. `n sin ^(n-1) x cos nx `
D. `n sin^(n-1) x cos (n+ 1) x`

Explanation

To differentiate the function \( y = \sin(nx) \cos(nx) \), we will use the product rule of differentiation. The product rule states that if \( y = u \cdot v \), then

\[ \frac{dy}{dx} = u'v + uv' \]

where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

### Step-by-Step Solution:

1. **Identify \( u \) and \( v \)**:
- Let \( u = \sin(nx) \)
- Let \( v = \cos(nx) \)

2. **Differentiate \( u \) and \( v \)**:
- To find \( u' \):
\[ u' = \frac{d}{dx}(\sin(nx)) = n \cos(nx) \]
- To find \( v' \):
\[ v' = \frac{d}{dx}(\cos(nx)) = -n \sin(nx) \]

3. **Apply the product rule**:
\[ \frac{dy}{dx} = u'v + uv' \]
Substituting \( u, u', v, \) and \( v' \):
\[ \frac{dy}{dx} = (n \cos(nx)) \cdot (\cos(nx)) + (\sin(nx)) \cdot (-n \sin(nx)) \]

4. **Simplify the expression**:
\[ \frac{dy}{dx} = n \cos^2(nx) - n \sin^2(nx) \]

5. **Factor out common terms**:
\[ \frac{dy}{dx} = n (\cos^2(nx) - \sin^2(nx)) \]

6. **Use the trigonometric identity**:
- Recall that \( \cos^2(a) - \sin^2(a) = \cos(2a) \):
\[ \frac{dy}{dx} = n \cos(2nx) \]

### Final Answer:
\[ \frac{dy}{dx} = n \cos(2nx) \]

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