Find the second derivatives of the following functions :
`e^(6x)cos3x`

Answer

To find the second derivative of the function \( y = e^{6x} \cos(3x) \), we will follow these steps:

### Step 1: Find the First Derivative
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:
- \( u = e^{6x} \) and \( v = \cos(3x) \)

Now, we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = 6e^{6x} \) (using the chain rule)
- \( \frac{dv}{dx} = -3\sin(3x) \) (using the chain rule)

Now, applying the product rule:
\[ \frac{dy}{dx} = e^{6x} \cdot (-3\sin(3x)) + \cos(3x) \cdot (6e^{6x}) \]
This simplifies to:
\[ \frac{dy}{dx} = -3e^{6x} \sin(3x) + 6e^{6x} \cos(3x) \]
We can factor out \( e^{6x} \):
\[ \frac{dy}{dx} = e^{6x} (6\cos(3x) - 3\sin(3x)) \]

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

Using the product rule again:
Let:
- \( u = e^{6x} \)
- \( v = 6\cos(3x) - 3\sin(3x) \)

We need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = 6e^{6x} \)
- For \( \frac{dv}{dx} \):
- \( \frac{d}{dx}(6\cos(3x)) = -18\sin(3x) \)
- \( \frac{d}{dx}(-3\sin(3x)) = -9\cos(3x) \)

Thus,
\[ \frac{dv}{dx} = -18\sin(3x) - 9\cos(3x) \]

Now, applying the product rule:
\[ \frac{d^2y}{dx^2} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values:
\[ \frac{d^2y}{dx^2} = e^{6x} (-18\sin(3x) - 9\cos(3x)) + (6\cos(3x) - 3\sin(3x))(6e^{6x}) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = e^{6x} (-18\sin(3x) - 9\cos(3x)) + 36e^{6x}\cos(3x) - 18e^{6x}\sin(3x) \]
Combining like terms:
\[ \frac{d^2y}{dx^2} = e^{6x} \left( (-18 - 18)\sin(3x) + (36 - 9)\cos(3x) \right) \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = e^{6x} (-36\sin(3x) + 27\cos(3x)) \]

### Final Answer
Thus, the second derivative of the function \( y = e^{6x} \cos(3x) \) is:
\[ \frac{d^2y}{dx^2} = e^{6x} (27\cos(3x) - 36\sin(3x)) \]