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How do the constant, sum, difference, and constant-multiple rules let you differentiate any polynomial term by term?

Topic 2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple: apply the basic linearity rules of differentiation to combine derivatives of individual terms.

A focused answer to AP Calculus AB Topic 2.6, covering the constant rule, constant-multiple rule, and sum and difference rules that let you differentiate polynomials term by term, with worked examples.

Generated by Claude Opus 4.88 min answer

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  1. What this topic is asking
  2. The four rules
  3. Differentiating term by term
  4. A common subtlety
  5. Why these rules hold
  6. Why linearity makes calculus tractable

What this topic is asking

The College Board (Topic 2.6) gives the linearity rules of differentiation: the derivative of a constant is zero, a constant factor pulls out, and the derivative of a sum or difference is the sum or difference of the derivatives. Together with the power rule, these let you differentiate any polynomial term by term.

The four rules

The constant rule reflects that a constant function has slope zero everywhere. The constant-multiple rule reflects that scaling a function scales its slope. The sum and difference rules reflect that derivatives add the way the functions do.

Differentiating term by term

A common subtlety

These rules cover sums, differences, and constant multiples - but not products or quotients of two variable functions. ddx[f(x)g(x)]\frac{d}{dx}[f(x)g(x)] is not f(x)g(x)f'(x)g'(x); products need the product rule (Topic 2.8) and quotients need the quotient rule (Topic 2.9). The linearity rules only split additive structure, not multiplicative structure.

Why these rules hold

Each linearity rule traces straight back to the limit definition of the derivative. The constant rule holds because a constant function never changes, so every difference quotient cch\frac{c - c}{h} is zero, and its limit is zero. The constant-multiple rule holds because a constant factor can be pulled outside a limit: limh0cf(x+h)cf(x)h=climh0f(x+h)f(x)h=cf(x)\lim_{h \to 0} \frac{c f(x+h) - c f(x)}{h} = c \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = c f'(x). The sum rule holds because the limit of a sum is the sum of the limits, so the difference quotient of f+gf + g splits into the difference quotients of ff and of gg separately. Knowing that these rules are consequences of limit laws you already met in Unit 1 - not new assumptions - ties the two units together and explains why they are so reliable.

Why linearity makes calculus tractable

Almost every function you differentiate is built from simpler pieces by addition and scaling. Linearity lets you break a complicated expression into manageable terms, differentiate each with a known rule, and reassemble. It is the reason the power rule plus these four rules already cover all polynomials and most early AP problems. The same linearity carries over to every later rule and to integration as well, so the term-by-term habit you build here pays off throughout the course. A typical exam item gives a polynomial or a sum of power and transcendental terms and asks for the derivative, the slope at a point, or the equation of a tangent line; in each case the work is to differentiate term by term and then substitute. Treating differentiation as a routine that distributes across sums, rather than something to be done all at once, is what keeps these problems fast and error-free.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice, no calculator). If f(x)=3x42x+7f(x) = 3x^4 - 2x + 7, then f(x)=f'(x) = (A) 12x3212x^3 - 2 (B) 12x32+712x^3 - 2 + 7 (C) 3x323x^3 - 2 (D) 12x32x12x^3 - 2x
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The correct answer is (A), 12x3212x^3 - 2.

Differentiate term by term: ddx[3x4]=12x3\frac{d}{dx}[3x^4] = 12x^3 (constant-multiple and power rules), ddx[2x]=2\frac{d}{dx}[-2x] = -2, and ddx[7]=0\frac{d}{dx}[7] = 0 (constant rule). Summing gives 12x3212x^3 - 2. The constant 77 disappears; choice (B) wrongly keeps it.

AP 2023 (style)3 marksSection II (free response, no calculator). Let f(x)=6x4x+5f(x) = 6\sqrt{x} - \frac{4}{x} + 5. (a) Rewrite ff using exponents. (b) Differentiate ff term by term, naming the rules used. (c) Evaluate f(4)f'(4).
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A 3-point linearity question.

(a) (1 point) f(x)=6x1/24x1+5f(x) = 6x^{1/2} - 4x^{-1} + 5.
(b) (1 point) By the constant-multiple and power rules, ddx[6x1/2]=3x1/2\frac{d}{dx}[6x^{1/2}] = 3x^{-1/2} and ddx[4x1]=4x2\frac{d}{dx}[-4x^{-1}] = 4x^{-2}; by the constant rule, ddx[5]=0\frac{d}{dx}[5] = 0. So f(x)=3x1/2+4x2=3x+4x2f'(x) = 3x^{-1/2} + 4x^{-2} = \frac{3}{\sqrt{x}} + \frac{4}{x^2}.
(c) (1 point) f(4)=34+416=32+14=74f'(4) = \frac{3}{\sqrt{4}} + \frac{4}{16} = \frac{3}{2} + \frac{1}{4} = \frac{7}{4}.

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