Find the second derivatives of the following functions :
`e^(x)sin5x`

Answer

To find the second derivative of the function \( y = e^x \sin(5x) \), we will follow these steps:

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

Using the product rule, which states that if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \), we can let:
- \( u = e^x \) and \( v = \sin(5x) \)

Now, we need to find \( u' \) and \( v' \):
- \( u' = \frac{d}{dx}(e^x) = e^x \)
- \( v' = \frac{d}{dx}(\sin(5x)) = 5\cos(5x) \)

Now we apply the product rule:
\[ \frac{dy}{dx} = u'v + uv' = e^x \sin(5x) + e^x (5\cos(5x)) \]
This simplifies to:
\[ \frac{dy}{dx} = e^x \sin(5x) + 5e^x \cos(5x) \]

### Step 2: Simplify the first derivative

We can factor out \( e^x \):
\[ \frac{dy}{dx} = e^x (\sin(5x) + 5\cos(5x)) \]

### Step 3: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \) again using the product rule:
Let \( u = e^x \) and \( v = \sin(5x) + 5\cos(5x) \).

We need to find \( u' \) and \( v' \):
- \( u' = e^x \)
- To find \( v' \):
- \( v' = \frac{d}{dx}(\sin(5x)) + 5\frac{d}{dx}(\cos(5x)) \)
- \( v' = 5\cos(5x) - 25\sin(5x) \)

Now we apply the product rule:
\[ \frac{d^2y}{dx^2} = u'v + uv' = e^x(\sin(5x) + 5\cos(5x)) + e^x(5\cos(5x) - 25\sin(5x)) \]

### Step 4: Combine like terms

Combining the terms:
\[ \frac{d^2y}{dx^2} = e^x(\sin(5x) + 5\cos(5x) + 5\cos(5x) - 25\sin(5x)) \]
\[ = e^x((-24\sin(5x) + 10\cos(5x))) \]

### Final Result

Thus, the second derivative of the function \( y = e^x \sin(5x) \) is:
\[ \frac{d^2y}{dx^2} = e^x(-24\sin(5x) + 10\cos(5x)) \]