If `y=e^(x) sin 3x ,then (d^(2)y)/(dx^(2))=`

Options

A. `e^(x) (8sin 3x - 6cos 3x )`
B. `e^(x) ( 6cos 3x- 8sin 3x )`
C. `e^(x) (3sin 3x - 4cos 3x )
D. `e^(x) ( 3cos 3x- 4sin 3x )`

Answer

To solve the problem \( y = e^x \sin(3x) \) and find the second derivative \( \frac{d^2y}{dx^2} \), we will follow these steps:

### Step 1: First Derivative
We will apply the product rule of differentiation, which states that if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \).

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

Now, we differentiate \( u \) and \( v \):
- \( u' = \frac{d}{dx}(e^x) = e^x \)
- \( v' = \frac{d}{dx}(\sin(3x)) = 3\cos(3x) \) (using the chain rule)

Now, applying the product rule:
\[ \frac{dy}{dx} = u'v + uv' = e^x \sin(3x) + e^x (3\cos(3x)) \]
\[ \frac{dy}{dx} = e^x \sin(3x) + 3e^x \cos(3x) \]
\[ \frac{dy}{dx} = e^x (\sin(3x) + 3\cos(3x)) \]

### Step 2: Second Derivative
Now we need to differentiate \( \frac{dy}{dx} \) again to find \( \frac{d^2y}{dx^2} \).

Using the product rule again:
Let \( u = e^x \) and \( v = \sin(3x) + 3\cos(3x) \).

Now, we differentiate \( v \):
- \( v' = \frac{d}{dx}(\sin(3x) + 3\cos(3x)) = 3\cos(3x) - 9\sin(3x) \)

Now, applying the product rule:
\[ \frac{d^2y}{dx^2} = u'v + uv' = e^x (\sin(3x) + 3\cos(3x)) + e^x (3\cos(3x) - 9\sin(3x)) \]
\[ \frac{d^2y}{dx^2} = e^x (\sin(3x) + 3\cos(3x) + 3\cos(3x) - 9\sin(3x)) \]
\[ \frac{d^2y}{dx^2} = e^x ((1 - 9)\sin(3x) + (3 + 3)\cos(3x)) \]
\[ \frac{d^2y}{dx^2} = e^x (-8\sin(3x) + 6\cos(3x)) \]

### Final Answer
Thus, the second derivative is:
\[ \frac{d^2y}{dx^2} = e^x (6\cos(3x) - 8\sin(3x)) \]