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AP® Calculus AB · Unit 2 · Topic 2.6

Derivative Rules: Constant, Sum, Difference and Constant Multiple: Revision Notes (Calculus AB 2.6)

One-page recap of the constant, constant multiple, sum and difference rules, how to differentiate polynomials term by term, and the slips that cost marks.

Short on time? This page is the recap. For the proofs from the limit definition, the graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.

Notation: f′(x), dy/dx and d/dx[…] all mean the derivative with respect to x. Here c and k are constants.

Recap

  • Each rule follows from the limit definition plus a limit property from Topic 1.5.
  • A constant term shifts the graph up or down. Steepness does not change, so its derivative is 0.
  • A coefficient stretches the graph vertically and multiplies every slope by the same factor.
  • Sums and differences can be differentiated one term at a time.
  • With the power rule, these four rules let you differentiate any polynomial and any sum of powers of x.
  • They do not split products or quotients.

Key relationships

Rule Statement Example
Constant d/dx[c] = 0 d/dx[π²] = 0
Constant multiple d/dx[k·f(x)] = k·f′(x) d/dx[7x³] = 21x²
Sum d/dx[f + g] = f′ + g′ d/dx[x⁴ + x⁻¹] = 4x³ − x⁻²
Difference d/dx[f − g] = f′ − g′ d/dx[x² − 5x] = 2x − 5
Linearity d/dx[a·f + b·g] = a·f′ + b·g′ f′(3) = 2, g′(3) = −5 → d/dx[4f − g] at 3 is 13

Rewrite first

  • Expand brackets: (2x − 3)² = 4x² − 12x + 9.
  • Split a fraction with a one-term denominator: (x³ + 5)/x = x² + 5x⁻¹.
  • Write roots and reciprocals as powers: 3/x² = 3x⁻², 5√x = 5x^(1/2).
  • A denominator with several terms needs the quotient rule (Topic 2.9).

Mistakes to avoid

  1. Keeping the constant term: d/dx[x² + 7] = 2x, not 2x + 7.
  2. Losing a coefficient: d/dx[5x] = 5, not 0.
  3. Differentiating π², e or √2 as if they contained x.
  4. Splitting products or quotients into products or quotients of derivatives.
  5. Differentiating (2x − 3)² as 2(2x − 3). Expand first.
  6. Sign slips: d/dx[−3x⁻²] = +6x⁻³.

Quick self-check

  1. Find d/dx[4x³ − 2x² + 9]. (12x² − 4x)
  2. Let f(x) = 6/x − 2√x. Find f′(1). (f′(x) = −6x⁻² − x^(−1/2), so f′(1) = −6 − 1 = −7)
  3. Find d/dx[(x + 2)(x − 5)]. (Expand: x² − 3x − 10, so the derivative is 2x − 3)
  4. Why do y = x³ and y = x³ − 40 have the same derivative? (The second is the first shifted down 40 units. A shift does not change any slope.)

Next: practice questions.

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