AP Calculus AB and BC

Derivative of x^3 e^x: Product Rule

The derivative of x cubed times e to the x is 3x squared e to the x plus x cubed e to the x. Factoring gives x squared times e to the x times 3 plus x, so the critical points are at x equals 0 and x equals negative 3.

ddx[x3ex]=x2ex(3+x)\frac{d}{dx}\left[x^{3}e^{x}\right] = x^{2}e^{x}(3+x)

The product rule, then factor

ddx(x3ex)=3x2ex+x3ex=x2ex(3+x)\frac{d}{dx}\left(x^{3}e^{x}\right) = 3x^{2}e^{x} + x^{3}e^{x} = x^{2}e^{x}(3+x)

Always factor after a product rule. The unfactored form hides the critical points; the factored form displays them.

Reading the critical points

Since exe^{x} is never zero, the derivative vanishes only where x2(3+x)=0x^{2}(3+x) = 0, at x=0x = 0 and x=−3x = -3.

At x=−3x = -3 the factor (3+x)(3+x) changes sign, so there is a minimum. At x=0x = 0 the factor x2x^{2} touches zero without changing sign, so there is NO extremum there, only a flat spot.

Common mistakes

  • Multiplying the derivatives to get 3x2ex3x^{2}e^{x} alone.
  • Assuming every zero of the derivative is an extremum. At x=0x = 0 the squared factor prevents a sign change.
  • Setting ex=0e^{x} = 0. It never is.

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^3 e^x?

It is 3x2ex+x3ex3x^{2}e^{x} + x^{3}e^{x}, or x2ex(3+x)x^{2}e^{x}(3+x) factored.

Where are the critical points?

At x=0x = 0 and x=−3x = -3. Only x=−3x = -3 is an extremum; at x=0x = 0 the derivative touches zero without changing sign.