Differentiate the following w.r.t. x :
`e^(-3x)sin^(2)3x`

Answer

To differentiate the function \( y = e^{-3x} \sin^2(3x) \) with respect to \( x \), 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:

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

Here, we can let:
- \( u = e^{-3x} \)
- \( v = \sin^2(3x) \)

### Step 1: Differentiate \( u \) and \( v \)

First, we need to find the derivatives of \( u \) and \( v \).

1. **Differentiate \( u \)**:
\[ u = e^{-3x} \implies \frac{du}{dx} = -3e^{-3x} \]

2. **Differentiate \( v \)**:
To differentiate \( v = \sin^2(3x) \), we will use the chain rule. Let \( w = \sin(3x) \), then \( v = w^2 \).
\[ \frac{dv}{dx} = 2w \frac{dw}{dx} = 2\sin(3x) \cdot \frac{d}{dx}(\sin(3x)) \]
Now, differentiate \( \sin(3x) \):
\[ \frac{d}{dx}(\sin(3x)) = 3\cos(3x) \]
Therefore,
\[ \frac{dv}{dx} = 2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x) \]

### Step 2: 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 found:
\[ \frac{dy}{dx} = e^{-3x} \cdot (6\sin(3x)\cos(3x)) + \sin^2(3x) \cdot (-3e^{-3x}) \]

### Step 3: Simplify the Expression

Now, we can simplify the expression:
\[ \frac{dy}{dx} = 6e^{-3x}\sin(3x)\cos(3x) - 3e^{-3x}\sin^2(3x) \]

### Step 4: Factor Out Common Terms

Notice that \( e^{-3x} \) is a common factor:
\[ \frac{dy}{dx} = e^{-3x} \left( 6\sin(3x)\cos(3x) - 3\sin^2(3x) \right) \]

### Final Answer

Thus, the derivative of \( y = e^{-3x} \sin^2(3x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = e^{-3x} \left( 6\sin(3x)\cos(3x) - 3\sin^2(3x) \right) \]