Differentiate `sinx sin 2x`

Answer

To differentiate the function \( y = \sin x \sin 2x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = \sin x \) and \( v = \sin 2x \).

2. **Differentiate Each Function**:
- The derivative of \( u \) with respect to \( x \) is:
\[ \frac{du}{dx} = \cos x \]
- The derivative of \( v \) with respect to \( x \) requires the chain rule. Since \( v = \sin(2x) \), we differentiate:
\[ \frac{dv}{dx} = \cos(2x) \cdot \frac{d}{dx}(2x) = \cos(2x) \cdot 2 = 2\cos(2x) \]

3. **Apply the Product Rule**:
Now we apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we have:
\[ \frac{dy}{dx} = \sin x \cdot (2\cos(2x)) + \sin(2x) \cdot \cos x \]

4. **Simplify the Expression**:
This gives us:
\[ \frac{dy}{dx} = 2\sin x \cos(2x) + \sin(2x) \cos x \]

### Final Answer:
Thus, the derivative of \( y = \sin x \sin 2x \) is:
\[ \frac{dy}{dx} = 2\sin x \cos(2x) + \sin(2x) \cos x \]