Find the second derivative of `e^(3x)sin4x.`

Answer

To find the second derivative of the function \( y = e^{3x} \sin(4x) \), we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

We will use the product rule for differentiation, which states that if \( y = u \cdot v \), then:

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

Here, let:
- \( u = e^{3x} \) and \( v = \sin(4x) \)

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):

1. \( \frac{du}{dx} = 3e^{3x} \) (using the chain rule)
2. \( \frac{dv}{dx} = 4\cos(4x) \) (using the chain rule)

Now applying the product rule:

\[ \frac{dy}{dx} = e^{3x} \cdot 4\cos(4x) + \sin(4x) \cdot 3e^{3x} \]

This simplifies to:

\[ \frac{dy}{dx} = 4e^{3x} \cos(4x) + 3e^{3x} \sin(4x) \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4e^{3x} \cos(4x) + 3e^{3x} \sin(4x)) \]

Again, we will use the product rule on both terms.

#### For the first term \( 4e^{3x} \cos(4x) \):

Let:
- \( u_1 = 4e^{3x} \) and \( v_1 = \cos(4x) \)

Then:

1. \( \frac{du_1}{dx} = 12e^{3x} \)
2. \( \frac{dv_1}{dx} = -4\sin(4x) \)

Applying the product rule:

\[ \frac{d}{dx}(4e^{3x} \cos(4x)) = 4e^{3x}(-4\sin(4x)) + \cos(4x)(12e^{3x}) \]

This simplifies to:

\[ = -16e^{3x} \sin(4x) + 12e^{3x} \cos(4x) \]

#### For the second term \( 3e^{3x} \sin(4x) \):

Let:
- \( u_2 = 3e^{3x} \) and \( v_2 = \sin(4x) \)

Then:

1. \( \frac{du_2}{dx} = 9e^{3x} \)
2. \( \frac{dv_2}{dx} = 4\cos(4x) \)

Applying the product rule:

\[ \frac{d}{dx}(3e^{3x} \sin(4x)) = 3e^{3x}(4\cos(4x)) + \sin(4x)(9e^{3x}) \]

This simplifies to:

\[ = 12e^{3x} \cos(4x) + 9e^{3x} \sin(4x) \]

### Step 3: Combine the derivatives

Now, we combine the results from both terms:

\[ \frac{d^2y}{dx^2} = (-16e^{3x} \sin(4x) + 12e^{3x} \cos(4x)) + (12e^{3x} \cos(4x) + 9e^{3x} \sin(4x)) \]

Combining like terms:

\[ = (-16e^{3x} \sin(4x) + 9e^{3x} \sin(4x)) + (12e^{3x} \cos(4x) + 12e^{3x} \cos(4x)) \]

This simplifies to:

\[ = -7e^{3x} \sin(4x) + 24e^{3x} \cos(4x) \]

### Final Answer

Thus, the second derivative is:

\[ \frac{d^2y}{dx^2} = 24e^{3x} \cos(4x) - 7e^{3x} \sin(4x) \]

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