The second order derivative if `y=e^(2x^2)` is:

Options

A. `4e^(2x^2)(4x^2+3)`
B. `4e^(2x^2)(4x^2-1)`
C. `4e^(2x^2)(4x^2+1)`
D. `e^(2x^2)(4x^2+1)`

Answer

To find the second order derivative of the function \( y = e^{2x^2} \), we will follow these steps:

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

Given:
\[ y = e^{2x^2} \]

To differentiate \( y \) with respect to \( x \), we will use the chain rule. The chain rule states that if \( y = e^{u} \), where \( u = 2x^2 \), then:
\[ \frac{dy}{dx} = e^{u} \cdot \frac{du}{dx} \]

First, we need to find \( \frac{du}{dx} \):
\[ u = 2x^2 \implies \frac{du}{dx} = 4x \]

Now, applying the chain rule:
\[ \frac{dy}{dx} = e^{2x^2} \cdot 4x = 4x e^{2x^2} \]

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

Now we differentiate \( \frac{dy}{dx} = 4x e^{2x^2} \) again to find the second derivative.

Using the product rule, where if \( f(x) = u(x)v(x) \), then:
\[ \frac{d}{dx}[u v] = u \frac{dv}{dx} + v \frac{du}{dx} \]

Let:
- \( u = 4x \)
- \( v = e^{2x^2} \)

Now we differentiate \( u \) and \( v \):
\[ \frac{du}{dx} = 4 \]
\[ \frac{dv}{dx} = e^{2x^2} \cdot \frac{d(2x^2)}{dx} = e^{2x^2} \cdot 4x \]

Now applying the product rule:
\[ \frac{d^2y}{dx^2} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values:
\[ \frac{d^2y}{dx^2} = 4x \cdot (4x e^{2x^2}) + e^{2x^2} \cdot 4 \]
\[ = 16x^2 e^{2x^2} + 4 e^{2x^2} \]

Now, we can factor out \( 4 e^{2x^2} \):
\[ \frac{d^2y}{dx^2} = 4 e^{2x^2} (4x^2 + 1) \]

### Final Answer
Thus, the second order derivative is:
\[ \frac{d^2y}{dx^2} = 4 e^{2x^2} (4x^2 + 1) \]

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