1. **Problem:** Find the derivative of $y = \sin(3x) \cos(2x)$.\n\n2. **Formula:** Use the product rule: if $y = u v$, then $y' = u' v + u v'$. Here, $u = \sin(3x)$ and $v = \cos(2x)$.\n\n3. **Derivatives:**\nCalculate $u'$ using the chain rule: $u' = \cos(3x) \cdot 3 = 3 \cos(3x)$.\nCalculate $v'$ using the chain rule: $v' = -\sin(2x) \cdot 2 = -2 \sin(2x)$.\n\n4. **Apply product rule:**\n$$y' = u' v + u v' = 3 \cos(3x) \cdot \cos(2x) + \sin(3x) \cdot (-2 \sin(2x)) = 3 \cos(3x) \cos(2x) - 2 \sin(3x) \sin(2x)$$\n\n5. **Final answer:**\n$$y' = 3 \cos(3x) \cos(2x) - 2 \sin(3x) \sin(2x)$$