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

Options

A. `2e^(x) ` sin x
B. `2e^(x) ` ( sin x - cos x )
C. `2e^(x) (cos x - sin ) `
D. `2e^(x) cos x`

Answer

To find the second derivative of the function \( y = e^x \sin x \), we will use the product rule for differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step 1: Differentiate \( y \)

Let:
- \( u = e^x \)
- \( v = \sin x \)

Now, we can find the first derivative \( \frac{dy}{dx} \):

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

Calculating \( \frac{dv}{dx} \) and \( \frac{du}{dx} \):

- \( \frac{du}{dx} = e^x \)
- \( \frac{dv}{dx} = \cos x \)

Substituting these into the product rule:

\[ \frac{dy}{dx} = e^x \cos x + \sin x e^x \]

This simplifies to:

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

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

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

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

Using the product rule again:

Let:
- \( u = e^x \)
- \( v = \cos x + \sin x \)

Now, we find \( \frac{d^2y}{dx^2} \):

\[ \frac{d^2y}{dx^2} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Calculating \( \frac{dv}{dx} \):

\[ \frac{dv}{dx} = -\sin x + \cos x \]

Now substituting back into the product rule:

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

This simplifies to:

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

Combining like terms:

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

### Final Result

Thus, the second derivative is:

\[ \frac{d^2y}{dx^2} = 2 e^x \cos x \]