Derivatives: rules and functions

Product rule

The product rule, (fg)′=f′g+fg′(fg)'=f'g+fg', differentiates a product of two functions.

(fg)′=f′g+fg′(fg)'=f'g+fg'

The product rule is: differentiate one factor at a time and add. Differentiate the first and keep the second, plus keep the first and differentiate the second.

(fg)′=f′g+fg′(fg)'=f'g+fg'

If you can expand first, expanding is sometimes faster.

Examples

  1. 01Differentiateddx ((x+1)(x2+2))\displaystyle \frac{d}{dx}\,((x+1)(x^{2}+2))AnswerHide
    Answer3x2+2x+2\displaystyle 3x^{2}+2x+2
  2. 02Differentiateddx (x2(x3−1))\displaystyle \frac{d}{dx}\,(x^{2}(x^{3}-1))AnswerHide
    Answer5x4−2x\displaystyle 5x^{4}-2x
  3. 03Differentiateddx ((2x−1)(x2+x))\displaystyle \frac{d}{dx}\,((2x-1)(x^{2}+x))AnswerHide
    Answer6x2+2x−1\displaystyle 6x^{2}+2x-1
  4. 04Differentiateddx x2sin⁡x\displaystyle \frac{d}{dx}\, x^{2}\sin xAnswerHide
    Answer2xsin⁡x+x2cos⁡x\displaystyle 2x\sin x + x^{2}\cos x
  5. 05Differentiateddx xex\displaystyle \frac{d}{dx}\, xe^{x}AnswerHide
    Answer(x+1)ex\displaystyle (x+1)e^{x}
  6. 06Differentiateddx x2log⁡x\displaystyle \frac{d}{dx}\, x^{2}\log xAnswerHide
    Answer2xlog⁡x+x\displaystyle 2x\log x + x

FAQ

Do the product rule and expanding first give the same result?
Yes. For two polynomials, expanding is often faster. When exe^x or sin⁡x\sin x is involved, you need the product rule.

Published 2026-09-03 · Updated 2026-09-05