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 differentiate the function \( y = \sin^n x \cos(nx) \), 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:

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

where \( u' \) is the derivative of \( u \) and \( v' \) is the derivative of \( v \).

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = \sin^n x \)
- \( v = \cos(nx) \)

### Step 2: Differentiate \( u \) and \( v \)
1. **Differentiate \( u \)**:
- Using the chain rule, we have:
\[ u' = n \sin^{n-1} x \cdot \cos x \]

2. **Differentiate \( v \)**:
- Using the chain rule, we have:
\[ v' = -\sin(nx) \cdot n \]

### Step 3: Apply the product rule
Now we can substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule formula:
\[ \frac{dy}{dx} = u'v + uv' \]
Substituting the values:
\[ \frac{dy}{dx} = (n \sin^{n-1} x \cos x) \cdot (\cos(nx)) + (\sin^n x) \cdot (-n \sin(nx)) \]

### Step 4: Simplify the expression
Now, let's simplify the expression:
\[ \frac{dy}{dx} = n \sin^{n-1} x \cos x \cos(nx) - n \sin^n x \sin(nx) \]

### Step 5: Factor out common terms
We can factor out \( n \) and \( \sin^{n-1} x \):
\[ \frac{dy}{dx} = n \sin^{n-1} x \left( \cos x \cos(nx) - \sin x \sin(nx) \right) \]

### Step 6: Use the cosine addition formula
Using the cosine addition formula \( \cos(a + b) = \cos a \cos b - \sin a \sin b \):
\[ \cos x \cos(nx) - \sin x \sin(nx) = \cos(x + nx) = \cos((n + 1)x) \]

### Final Result
Thus, we can write:
\[ \frac{dy}{dx} = n \sin^{n-1} x \cos((n + 1)x) \]