Differentiate `sin^(m)x*cos^(n)x`

Explanation

To differentiate the function \( y = \sin^m x \cdot \cos^n x \), we will use the product rule of differentiation. The product rule states that if \( y = u \cdot v \), then the derivative \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).

### Step-by-Step Solution:

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

2. **Differentiate \( u \)**:
To differentiate \( u = \sin^m x \), we use the chain rule:
\[ \frac{du}{dx} = m \sin^{m-1} x \cdot \cos x \]
(Here, we differentiate \( \sin x \) to get \( \cos x \)).

3. **Differentiate \( v \)**:
To differentiate \( v = \cos^n x \), we again use the chain rule:
\[ \frac{dv}{dx} = n \cos^{n-1} x \cdot (-\sin x) = -n \cos^{n-1} x \cdot \sin x \]
(Here, we differentiate \( \cos x \) to get \( -\sin x \)).

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

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

6. **Factor out common terms**:
We can factor out \( \sin^{m-1} x \cdot \cos^{n-1} x \):
\[ \frac{dy}{dx} = \sin^{m-1} x \cdot \cos^{n-1} x \left( -n \sin^2 x + m \cos^2 x \right) \]

### Final Result:
Thus, the derivative of \( y = \sin^m x \cdot \cos^n x \) is:
\[ \frac{dy}{dx} = \sin^{m-1} x \cdot \cos^{n-1} x \left( m \cos^2 x - n \sin^2 x \right) \]