Derivatives: rules and functions

Product rule

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

(fg)=fg+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)=fg+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(x31))\displaystyle \frac{d}{dx}\,(x^{2}(x^{3}-1))AnswerHide
    Answer5x42x\displaystyle 5x^{4}-2x
  3. 03Differentiateddx((2x1)(x2+x))\displaystyle \frac{d}{dx}\,((2x-1)(x^{2}+x))AnswerHide
    Answer6x2+2x1\displaystyle 6x^{2}+2x-1
  4. 04Differentiateddxx2sinx\displaystyle \frac{d}{dx}\, x^{2}\sin xAnswerHide
    Answer2xsinx+x2cosx\displaystyle 2x\sin x + x^{2}\cos x
  5. 05Differentiateddxxex\displaystyle \frac{d}{dx}\, xe^{x}AnswerHide
    Answer(x+1)ex\displaystyle (x+1)e^{x}
  6. 06Differentiateddxx2logx\displaystyle \frac{d}{dx}\, x^{2}\log xAnswerHide
    Answer2xlogx+x\displaystyle 2x\log x + x

FAQ

Q1Do the product rule and expanding first give the same result?Show answerHide
A
Yes. For two polynomials, expanding is often faster. When exe^x or sinx\sin x is involved, you need the product rule.

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Product rule

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