Derivative of xe^x: Proof, Step by Step Solution

The derivative of xe^x is (x + 1)e^x, or e^x + xe^x before you factor.

Searching “derivative of xe x” usually means homework or a test. So here’s the answer first, then the working, then the places where marks get lost.

☰ Table of Contents

Why You Need the Product Rule

xe^x is x times e^x, two functions multiplied. Differentiating each one and multiplying the results is the classic wrong move. It gives e^x. Try x = 1: that says the slope is about 2.72, when the real slope is about 5.44. A whole term has gone missing.

The product rule puts it back. If y = u·v, then y′ = u′v + uv′.

Solving It

Let u = x and v = e^x. So u′ = 1 and v′ = e^x.

y′ = (1)(e^x) + (x)(e^x) = e^x + xe^x = e^x(1 + x)

That’s the derivative of xe^x: (x + 1)e^x.

(Why does e^x stay the same when differentiated? e is the base where a^x has slope 1 at x = 0. From there the slope at any point equals the height of the curve.)

From First Principle

If your syllabus wants the limit definition, here it is.

f′(x) = lim (h → 0) [f(x + h) − f(x)] / h, with f(x) = xe^x

= lim [(x + h)e^(x + h) − xe^x] / h

Split e^(x + h) into e^x · e^h and take out e^x:

= lim e^x [(x + h)e^h − x] / h = lim e^x [xe^h + he^h − x] / h = lim e^x [x(e^h − 1) + he^h] / h = lim e^x [x(e^h − 1)/h + e^h]

Now use two standard limits: (e^h − 1)/h → 1 and e^h → 1, as h → 0.

f′(x) = e^x (x + 1)

A well known reference page ends this proof with “e^x + e^x”. That’s a typo. The line should be e^x + xe^x.

Checking the Answer

Logarithmic differentiation gives a second route for x > 0. If y = xe^x, then ln y = ln x + x. Differentiating, y′/y = 1/x + 1, and multiplying by xe^x leaves e^x + xe^x.

A numerical check works too. At x = 1 the formula says 2e ≈ 5.4366. The slope over a small gap comes to 5.863 for h = 0.1, 5.478 for h = 0.01 and 5.441 for h = 0.001. The values close in on 5.4366.

Higher Derivatives

The second derivative is (x + 2)e^x, the third is (x + 3)e^x, the fourth is (x + 4)e^x. The number next to x goes up by 1 each time, so the nth derivative is (x + n)e^x. The tenth is (x + 10)e^x.

Shape of the Graph

The slope is 1 at x = 0, about 5.44 at x = 1 and about 22.17 at x = 2. Set (x + 1)e^x = 0 and you get x = −1, because e^x is never zero. That’s the lowest point, where xe^x = −1/e ≈ −0.37. To the left of it the slope is negative, to the right it’s positive. Far to the left the slope heads toward 0.

Examples

Slope of xe^x at x = 2: (x + 1)e^x gives 3e² ≈ 22.17.

x²e^x: with u = x² and v = e^x, you get 2xe^x + x²e^x = (x² + 2x)e^x.

xe^(2x): the chain rule gives e^(2x) its own factor of 2, so the answer is e^(2x) + 2xe^(2x) = (2x + 1)e^(2x).

xe^(−x): the derivative of e^(−x) is −e^(−x), so you get e^(−x) − xe^(−x) = (1 − x)e^(−x).

For a position s(t) = te^t, the velocity is (t + 1)e^t and the acceleration is (t + 2)e^t. At t = 1 those are 2e ≈ 5.44 and 3e ≈ 8.15.

The integral goes the other way: ∫ xe^x dx = (x − 1)e^x + C. Differentiate it to check.

Mistakes to Avoid

Differentiating only one factor. Forgetting that e^x is its own derivative. Dropping the 2 in e^(2x). Stopping at e^x + xe^x when the question wants it factored. And confusing xe^x with e^(x²), which is a composite and has derivative 2xe^(x²).

It also helps to write u, v, u′ and v′ at the top of your working. If you slip later, the method marks are still yours.

Practice

Differentiate xe^x.
Find the slope of xe^x at x = 1.
Find the third derivative of xe^x.
Differentiate x³e^x.

Answers: (x + 1)e^x; 2e; (x + 3)e^x; (x³ + 3x²)e^x.

Conclusion

The derivative of xe^x is (x + 1)e^x, and the nth derivative is (x + n)e^x.

Frequently Asked Questions

What is the derivative of xe^x?

(x + 1)e^x.

Which rule is used?

The product rule.

What is it at x = 0?

What is the second derivative?

(x + 2)e^x.

Is xe^x the same as x(e^x)?

Yes, so the derivative is the same.