Find `dy/dx if y= sinx.cosx`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \sin x \cdot \cos x\), we will use the product rule of differentiation. The product rule states that if \(y = u \cdot v\), then:

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

where \(u = \sin x\) and \(v = \cos x\).

### Step-by-Step Solution:

1. **Identify \(u\) and \(v\)**:
\[ u = \sin x, \quad v = \cos x \]

2. **Differentiate \(u\) and \(v\)**:
- The derivative of \(u\) with respect to \(x\) is:
\[ \frac{du}{dx} = \cos x \]
- The derivative of \(v\) with respect to \(x\) is:
\[ \frac{dv}{dx} = -\sin x \]

3. **Apply the product rule**:
Using the product rule:
\[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \]
Substituting the values of \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\):
\[ \frac{dy}{dx} = \sin x \cdot (-\sin x) + \cos x \cdot \cos x \]

4. **Simplify the expression**:
\[ \frac{dy}{dx} = -\sin^2 x + \cos^2 x \]

5. **Use the Pythagorean identity**:
We know from trigonometric identities that:
\[ \cos^2 x - \sin^2 x = \cos(2x) \]
Therefore, we can rewrite the expression:
\[ \frac{dy}{dx} = \cos^2 x - \sin^2 x = \cos(2x) \]

### Final Result:
\[ \frac{dy}{dx} = \cos(2x) \]