If `y=sin2x.cos3x`, then `(dy)/(dx)`.

Answer

To find the derivative of \( y = \sin(2x) \cos(3x) \), 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:

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

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = \sin(2x) \) and \( v = \cos(3x) \).

2. **Differentiate \( u \)**:
To find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{d}{dx}(\sin(2x)) = \cos(2x) \cdot \frac{d}{dx}(2x) = 2\cos(2x) \]

3. **Differentiate \( v \)**:
To find \( \frac{dv}{dx} \):
\[ \frac{dv}{dx} = \frac{d}{dx}(\cos(3x)) = -\sin(3x) \cdot \frac{d}{dx}(3x) = -3\sin(3x) \]

4. **Apply the product rule**:
Now, substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule formula:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values:
\[ \frac{dy}{dx} = \sin(2x)(-3\sin(3x)) + \cos(3x)(2\cos(2x)) \]

5. **Simplify the expression**:
\[ \frac{dy}{dx} = -3\sin(2x)\sin(3x) + 2\cos(2x)\cos(3x) \]

6. **Final answer**:
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 2\cos(2x)\cos(3x) - 3\sin(2x)\sin(3x) \]