Derivative of x^2 e^x: Answer, Product Rule, Mistakes

How to differentiate x^2 e^x

This is a product of $x^{2}$ and $e^{x}$, so the product rule applies: differentiate each factor once while holding the other fixed, then add.

Take $f(x) = x^{2}$ and $g(x) = e^{x}$, so $f'(x) = 2x$ and $g'(x) = e^{x}$, since the exponential is its own derivative.

Both terms share a factor of $e^{x}$, so factor it out for the cleanest form.

What the factored form tells you

Written as $e^{x}(x^{2} + 2x) = e^{x}\cdot x(x+2)$, the derivative is $0$ only where $x(x+2) = 0$, since $e^{x}$ is never zero. So $x^{2}e^{x}$ has two critical points, at $x = 0$ and $x = -2$.

The reverse operation is integration by parts applied twice: $\int x^{2}e^{x}\,dx = e^{x}(x^{2} - 2x + 2) + C$. Note the derivative and the antiderivative are different polynomials in front of $e^{x}$.

Where the derivative of x^2 e^x shows up on the AP exam

The product rule is Topic 2.8, on both AB and BC, and a polynomial times $e^{x}$ is a favorite pairing because $e^{x}$ differentiates to itself, so the factoring step stays clean.

Common mistakes with the derivative of x^2 e^x

• Answering $2xe^{x}$, multiplying the derivatives $2x$ and $e^{x}$ instead of using the product rule.
• Answering $2xe^{x}$ by keeping only the first term and dropping $x^{2}e^{x}$.
• Answering $e^{x}(x^{2} - 2x + 2)$, which is the antiderivative, not the derivative.