Find the derivatives of the following :
`sinxsin2x`

Answer

To find the derivative of the function \( y = \sin x \sin 2x \), we can 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:

\[ (uv)' = u'v + uv' \]

### Step-by-step solution:

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

2. **Differentiate \( u \) and \( v \)**:
- The derivative of \( u \):
\[ u' = \frac{d}{dx}(\sin x) = \cos x \]
- The derivative of \( v \):
\[ v' = \frac{d}{dx}(\sin 2x) = 2 \cos 2x \]
(using the chain rule, where the derivative of \( \sin kx \) is \( k \cos kx \)).

3. **Apply the product rule**:
Now, apply the product rule:
\[ y' = u'v + uv' \]
Substituting the values we found:
\[ y' = (\cos x)(\sin 2x) + (\sin x)(2 \cos 2x) \]

4. **Simplify the expression**:
We can factor out common terms:
\[ y' = \cos x \sin 2x + 2 \sin x \cos 2x \]

5. **Final expression**:
Thus, the derivative of \( y = \sin x \sin 2x \) is:
\[ y' = \cos x \sin 2x + 2 \sin x \cos 2x \]