Question

For the function $f(x) = x^2e^{x^2}$, what is $f''(x)$?

• A. $e^{x^2} (2x^4 + 5x^2 + 1)$
• B. $2e^{x^2} (2x^4 + 5x^2 + 1)$
• C. $2e^{x^2} (2x^4 - 5x^2 + 1)$
• D. $e^{x^2} (2x^4 - 5x^2 + 1)$

Answer 1

Answer: (B)

2e^{x^2} (2x^4 + 5x^2 + 1)

Answer 2

To find the second derivative $f''(x)$ of the function $f(x) = x^2e^{x^2}$, we need to apply differentiation rules twice.

Step 1: Find the first derivative, $f'(x)$

The function $f(x) = x^2e^{x^2}$ is a product of two functions: $u(x) = x^2$ and $v(x) = e^{x^2}$. We use the product rule: $(uv)' = u'v + uv'$.

First, find the derivatives of $u(x)$ and $v(x)$:

• $u(x) = x^2 \implies u'(x) = 2x$
• $v(x) = e^{x^2}$. To differentiate $e^{x^2}$, we use the chain rule. Let $g(x) = x^2$, so $v(x) = e^{g(x)}$. Then $v'(x) = e^{g(x)} \cdot g'(x) = e^{x^2} \cdot (2x) = 2xe^{x^2}$.

Now, apply the product rule to find $f'(x)$:

$f'(x) = u'(x)v(x) + u(x)v'(x)$

$f'(x) = (2x)(e^{x^2}) + (x^2)(2xe^{x^2})$

$f'(x) = 2xe^{x^2} + 2x^3e^{x^2}$

Factor out $2xe^{x^2}$:

$f'(x) = 2xe^{x^2}(1 + x^2)$

Step 2: Find the second derivative, $f''(x)$

Now we need to differentiate $f'(x) = 2xe^{x^2}(1 + x^2)$. Again, this is a product of two functions. Let's rewrite $f'(x)$ as $f'(x) = (2x + 2x^3)e^{x^2}$.

Let $A(x) = 2x + 2x^3$ and $B(x) = e^{x^2}$. We use the product rule: $(AB)' = A'B + AB'$.

First, find the derivatives of $A(x)$ and $B(x)$:

• $A(x) = 2x + 2x^3 \implies A'(x) = \frac{d}{dx}(2x) + \frac{d}{dx}(2x^3) = 2 + 6x^2$
• $B(x) = e^{x^2}$. As calculated before, $B'(x) = 2xe^{x^2}$.

Now, apply the product rule to find $f''(x)$:

$f''(x) = A'(x)B(x) + A(x)B'(x)$

$f''(x) = (2 + 6x^2)(e^{x^2}) + (2x + 2x^3)(2xe^{x^2})$

$f''(x) = (2 + 6x^2)e^{x^2} + (4x^2 + 4x^4)e^{x^2}$

Factor out $e^{x^2}$:

$f''(x) = e^{x^2} [(2 + 6x^2) + (4x^2 + 4x^4)]$

Combine like terms inside the bracket:

$f''(x) = e^{x^2} [4x^4 + 6x^2 + 4x^2 + 2]$

$f''(x) = e^{x^2} [4x^4 + 10x^2 + 2]$

Factor out 2 from the bracket:

$f''(x) = 2e^{x^2} [2x^4 + 5x^2 + 1]$