Question

If f(x) = x²eˣ, what is f''(x), the second derivative of f(x)?

• A. (2x + 1)eˣ
• B. (x² + 2x + 2)eˣ
• C. (x² + 2x)eˣ
• D. (x² + 4x + 2)eˣ

Answer 1

Answer: (D)

(x² + 4x + 2)eˣ

Answer 2

To find the second derivative $f''(x)$ of the function $f(x) = x^2e^x$, we will follow these steps:

1. Find the first derivative, $f'(x)$.
2. Find the second derivative, $f''(x)$, by differentiating $f'(x)$.

We will use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

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

Given $f(x) = x^2e^x$.
Let $u(x) = x^2$ and $v(x) = e^x$.
Now, find the derivatives of $u(x)$ and $v(x)$:
$u'(x) = \frac{d}{dx}(x^2) = 2x$
$v'(x) = \frac{d}{dx}(e^x) = e^x$

Apply the product rule to find $f'(x)$:
$f'(x) = u'(x)v(x) + u(x)v'(x)$
$f'(x) = (2x)(e^x) + (x^2)(e^x)$
Factor out $e^x$:
$f'(x) = (2x + x^2)e^x$

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

Now we need to differentiate $f'(x) = (2x + x^2)e^x$.
Let $A(x) = 2x + x^2$ and $B(x) = e^x$.
Find the derivatives of $A(x)$ and $B(x)$:
$A'(x) = \frac{d}{dx}(2x + x^2) = 2 + 2x$
$B'(x) = \frac{d}{dx}(e^x) = e^x$

Apply the product rule again to find $f''(x)$:
$f''(x) = A'(x)B(x) + A(x)B'(x)$
$f''(x) = (2 + 2x)(e^x) + (2x + x^2)(e^x)$
Factor out $e^x$:
$f''(x) = e^x [(2 + 2x) + (2x + x^2)]$
Combine like terms inside the bracket:
$f''(x) = e^x [x^2 + 2x + 2x + 2]$
$f''(x) = e^x [x^2 + 4x + 2]$

So, $f''(x) = (x^2 + 4x + 2)e^x$.