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

Options

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

Answer

To find the derivative of \( y = \sin^n x \cdot \cos(nx) \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = \sin^n x \) and \( v = \cos(nx) \).

2. **Differentiate \( u \)**:
To differentiate \( u = \sin^n x \), we apply the chain rule:
\[ \frac{du}{dx} = n \sin^{n-1} x \cdot \frac{d}{dx}(\sin x) = n \sin^{n-1} x \cdot \cos x \]

3. **Differentiate \( v \)**:
To differentiate \( v = \cos(nx) \), we again apply the chain rule:
\[ \frac{dv}{dx} = -\sin(nx) \cdot \frac{d}{dx}(nx) = -n \sin(nx) \]

4. **Apply the product rule**:
Now, substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
\[ \frac{dy}{dx} = \sin^n x \cdot (-n \sin(nx)) + \cos(nx) \cdot (n \sin^{n-1} x \cdot \cos x) \]

5. **Simplify the expression**:
\[ \frac{dy}{dx} = -n \sin^n x \sin(nx) + n \sin^{n-1} x \cos(nx) \cos x \]

6. **Factor out common terms**:
\[ \frac{dy}{dx} = n \sin^{n-1} x \left( \cos(nx) \cos x - \sin^n x \sin(nx) \right) \]

7. **Use the product-to-sum identities**:
The expression \( \cos(nx) \cos x - \sin(nx) \sin x \) can be simplified using the cosine addition formula:
\[ \cos(nx + x) = \cos((n+1)x) \]
Therefore, we have:
\[ \frac{dy}{dx} = n \sin^{n-1} x \cos((n+1)x) \]

### Final Answer:
\[ \frac{dy}{dx} = n \sin^{n-1} x \cos((n+1)x) \]