AP Calculus AB and BC

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

Q: What is the derivative of x e^x?

\frac{d}{dx}(xe^{x}) = e^{x}(x+1). The product rule gives 1\cdot e^{x} + x\cdot e^{x} = e^{x} + xe^{x}, and factoring out e^{x} yields e^{x}(x+1).

Q: Why does e^x factor out?

Both terms of e^{x} + xe^{x} contain e^{x}, so it is a common factor. Pulling it out gives e^{x}(1 + x), the standard simplified form and the easiest one for finding critical points.

Q: Where is x e^x minimized?

Set the derivative e^{x}(x+1) to 0. Since e^{x} \ne 0, this needs x = -1, where the function reaches its minimum value -e^{-1} \approx -0.368.

The derivative of x times e^x with respect to x is e^x(x + 1). The product rule gives 1 times e^x plus x times e^x, and factoring out the common e^x leaves e^x(x + 1). The answer is not e^x, which would come from wrongly multiplying the two derivatives.

\frac{d}{dx}\left[xe^x\right] = e^x(x+1)

How to differentiate x e^x

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

\frac{d}{dx}\left[f(x)g(x)\right] = f'(x)g(x) + f(x)g'(x)

Take $f(x) = x$ and $g(x) = e^{x}$ , so $f'(x) = 1$ and $g'(x) = e^{x}$ , since the exponential is its own derivative.

\frac{d}{dx}\left(xe^{x}\right) = 1\cdot e^{x} + x\cdot e^{x}

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

e^{x} + xe^{x} = e^{x}(x+1)

What the factored form tells you

Written as $e^{x}(x+1)$ , the derivative is $0$ only where $x + 1 = 0$ , since $e^{x}$ is never zero. So $xe^{x}$ has exactly one critical point, at $x = -1$ .

The derivative is negative for $x < -1$ and positive for $x > -1$ , so the function falls then rises and $x = -1$ is a minimum. That minimum value is $-e^{-1}$ , a number that turns up often in optimization problems.

The reverse operation is integration by parts: $\int xe^{x}\,dx = e^{x}(x-1) + C$ , closely related to the derivative but with a minus inside.

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

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

Common mistakes with the derivative of x e^x

Answering $e^{x}$ , multiplying the derivatives $1$ and $e^{x}$ instead of using the product rule.
Answering $xe^{x}$ and keeping only the second term, dropping the $1\cdot e^{x}$ .
Answering $e^{x}(x-1)$ , which is the antiderivative, not the derivative. The derivative factors to $e^{x}(x+1)$ .

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of x e^x?

$\frac{d}{dx}(xe^{x}) = e^{x}(x+1)$ . The product rule gives $1\cdot e^{x} + x\cdot e^{x} = e^{x} + xe^{x}$ , and factoring out $e^{x}$ yields $e^{x}(x+1)$ .

Why does e^x factor out?

Both terms of $e^{x} + xe^{x}$ contain $e^{x}$ , so it is a common factor. Pulling it out gives $e^{x}(1 + x)$ , the standard simplified form and the easiest one for finding critical points.

Where is x e^x minimized?

Set the derivative $e^{x}(x+1)$ to $0$ . Since $e^{x} \ne 0$ , this needs $x = -1$ , where the function reaches its minimum value $-e^{-1} \approx -0.368$ .