If `y=e^(x^(2))sin2x`, then what is `(dy)/(dx)` at `x=pi` equal to?

Options

A. `(1+pi)e^(pi^(2))`
B. `2pie^(pi^(2))`
C. `2e^(pi^(2))`
D. `e^(pi^(2))`

Answer

To find the derivative of the function \( y = e^{x^2} \sin(2x) \) and evaluate it at \( x = \pi \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = u'v + uv' \]

### Step 1: Identify the functions
Let:
- \( u = e^{x^2} \)
- \( v = \sin(2x) \)

### Step 2: Find the derivatives \( u' \) and \( v' \)
1. **Finding \( u' \)**:
\[ u = e^{x^2} \implies u' = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x \]

2. **Finding \( v' \)**:
\[ v = \sin(2x) \implies v' = \cos(2x) \cdot \frac{d}{dx}(2x) = \cos(2x) \cdot 2 = 2\cos(2x) \]

### Step 3: Apply the product rule
Now, substituting \( u, v, u', \) and \( v' \) into the product rule:

\[ \frac{dy}{dx} = u'v + uv' = (e^{x^2} \cdot 2x) \sin(2x) + e^{x^2} (2\cos(2x)) \]

### Step 4: Simplify the expression
\[ \frac{dy}{dx} = 2x e^{x^2} \sin(2x) + 2 e^{x^2} \cos(2x) \]

### Step 5: Factor out common terms
\[ \frac{dy}{dx} = 2 e^{x^2} (x \sin(2x) + \cos(2x)) \]

### Step 6: Evaluate at \( x = \pi \)
Now, we need to evaluate \( \frac{dy}{dx} \) at \( x = \pi \):

\[ \frac{dy}{dx} \bigg|_{x = \pi} = 2 e^{\pi^2} \left( \pi \sin(2\pi) + \cos(2\pi) \right) \]

### Step 7: Calculate the trigonometric values
- \( \sin(2\pi) = 0 \)
- \( \cos(2\pi) = 1 \)

Substituting these values:

\[ \frac{dy}{dx} \bigg|_{x = \pi} = 2 e^{\pi^2} \left( \pi \cdot 0 + 1 \right) = 2 e^{\pi^2} \cdot 1 = 2 e^{\pi^2} \]

### Final Answer
Thus, the value of \( \frac{dy}{dx} \) at \( x = \pi \) is:

\[ \frac{dy}{dx} \bigg|_{x = \pi} = 2 e^{\pi^2} \]