derivative of `e^(-2x)sin 4 x`

Answer

To find the derivative of the function \( y = e^{-2x} \sin(4x) \), 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 \( uv \) is given by:

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

### Step-by-Step Solution:

1. **Identify \( u \) and \( v \)**:
Let \( u = e^{-2x} \) and \( v = \sin(4x) \).

2. **Find \( \frac{du}{dx} \)**:
To differentiate \( u = e^{-2x} \), we apply the chain rule:
\[ \frac{du}{dx} = e^{-2x} \cdot (-2) = -2e^{-2x} \]

3. **Find \( \frac{dv}{dx} \)**:
To differentiate \( v = \sin(4x) \), we again apply the chain rule:
\[ \frac{dv}{dx} = \cos(4x) \cdot \frac{d}{dx}(4x) = \cos(4x) \cdot 4 = 4\cos(4x) \]

4. **Apply the product rule**:
Now, we can 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 we found:
\[ \frac{dy}{dx} = e^{-2x} \cdot (4\cos(4x)) + \sin(4x) \cdot (-2e^{-2x}) \]

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

6. **Factor out common terms**:
We can factor out \( e^{-2x} \):
\[ \frac{dy}{dx} = e^{-2x} (4\cos(4x) - 2\sin(4x)) \]

### Final Answer:
Thus, the derivative of \( e^{-2x} \sin(4x) \) is:
\[ \frac{dy}{dx} = e^{-2x} (4\cos(4x) - 2\sin(4x)) \]