Unit 2Basic Rules
Concept 4.140Built for May 2027 AP Exam

Basic Rules

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Three rules do most of the work in this unit, and none of them requires a limit once the power rule is in hand. The derivative of a constant is zero, because a horizontal line has no slope. A constant multiple passes through the derivative unchanged, because scaling the output scales every slope by the same factor. A sum or difference differentiates term by term, because the limit of a sum is the sum of the limits. Those three, plus the power rule, differentiate every polynomial in one line. What they do not do is equally important: there is no rule that lets you differentiate a product, a quotient, or a composition by handling the pieces separately, and those three cases arrive in the next sections.

On the AP Exam

What you’ll see
A polynomial, a sum of a polynomial and a constant, or a term with a coefficient, inside a rate, a tangent, or a second derivative.
What you’ll do
Differentiate term by term, keep every coefficient, and let every standalone constant disappear.
Memory hook
Constants out, constants gone. A multiplier survives the derivative; an additive constant does not.
Use it on the exam
For MCQs: The distractors usually keep an additive constant or drop a coefficient. Count the terms in your derivative against the terms in the original: only the constant term should vanish. For FRQs: Speed matters here, because the derivative is a step rather than an answer. Write the derivative in one pass, then spend the space on the substitution and the interpretation the part is paying for.

Concept Review

Three Rules in One Pass

Figure 1 — Each card states one rule and applies it immediately. The rules act on single blocks, which is why a polynomial can be differentiated left to right without regrouping anything.

Three stacked rule cards for the constant rule, the constant multiple rule, and the sum and difference rule, each paired with a worked example and a one-line note

Why They Hold

Each rule is a limit statement, and each follows from the limit laws:

ddx[c]=lim⁡h→0c−ch=0,ddx[cf]=clim⁡h→0f(x+h)−f(x)h=cf′(x)\frac{d}{dx}[c] = \lim_{h\to 0}\frac{c-c}{h} = 0, \qquad \frac{d}{dx}[cf] = c\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = cf'(x)

ddx[f±g]=lim⁡h→0[f(x+h)−f(x)]±[g(x+h)−g(x)]h=f′(x)±g′(x)\frac{d}{dx}[f \pm g] = \lim_{h\to 0}\frac{[f(x+h)-f(x)] \pm [g(x+h)-g(x)]}{h} = f'(x) \pm g'(x)

Nothing in these arguments is specific to polynomials. They hold for every differentiable function, which is why the rules keep working when the pieces become trigonometric or exponential.

What the Rules Do Not Cover

You might hopeThe truthWhere it is handled
(fg)′=f′g′(fg)' = f'g'False in generalProduct rule
(fg)′=f′g′\left(\frac{f}{g}\right)' = \frac{f'}{g'}False in generalQuotient rule
(f∘g)′=f′∘g′(f \circ g)' = f' \circ g'False in generalChain rule

A single counterexample settles each row. With f(x)=xf(x) = x and g(x)=xg(x) = x, the product is x2x^2, whose derivative is 2x2x, while f′g′=1f'g' = 1. The failure is not a technicality; it is the reason the next three sections exist.

Reading a Derivative Backwards

Because coefficients pass through, you can often reconstruct the original from the derivative. If f′(x)=6x2−4f'(x) = 6x^2 - 4, then ff must contain 2x3−4x2x^3 - 4x plus some constant, since differentiating 2x32x^3 gives 6x26x^2 and differentiating −4x-4x gives −4-4. That reconstruction is antidifferentiation, and it works for exactly the same reason the rules do.

Common Misconception Check

  • Myth 1: "The derivative of 5x35x^3 is 5x25x^2." Truth: The power rule still multiplies by the exponent, so the answer is 15x215x^2. The constant multiple only preserves the factor of 5.
  • Myth 2: "A constant term becomes 1." Truth: It becomes 0. The value 1 is what the derivative of xx gives, not what the derivative of 77 gives.
  • Myth 3: "Sum rules let you split a product." Truth: Sums split; products do not. Confusing the two produces the most common structural error in this unit.

Topic-Specific Study Method

One Pass, Left to Right

Borrowed idea: Work from left to right, and treat each block as though the others were absent. A coefficient stays exactly where it is while its block is differentiated. A constant standing alone disappears, since it adds nothing to any slope. A subtraction sign belongs to the block that follows it, and a dropped sign stays invisible until the substitution step. When the expression is long, cover the other terms and differentiate what you can see. Then count the terms in your answer. Only the constant should have gone missing, and any other gap means a block was skipped.

GivenDerivativeWhat the rules did
7700Constant rule
7x27x^214x14xMultiple plus power
7x2−77x^2 - 714x14xConstant rule removed the tail
x34\frac{x^3}{4}3x24\frac{3x^2}{4}Multiple 14\frac{1}{4} passed through
2x−x22x - x^22−2x2 - 2xDifference, sign attached to the second block

① Differentiate five polynomials with a constant term, then count the terms in each answer. The count should drop by exactly one, and that check catches most slips faster than re-deriving.

② Write a function whose derivative is 4x3−6x+14x^3 - 6x + 1, then check it by differentiating back. Reconstructing forces you to see the rules in reverse.

③ Take a product such as x⋅xx \cdot x and differentiate it two ways, once correctly and once by the false rule (fg)′=f′g′(fg)' = f'g'. The difference between 2x2x and 11 is the reason the next section exists.

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate f(x)=3x4−2x2+5x−9f(x) = 3x^4 - 2x^2 + 5x - 9.

Standard Execution:

  1. Term by term. 3x4→12x33x^4 \to 12x^3, −2x2→−4x-2x^2 \to -4x, 5x→55x \to 5, −9→0-9 \to 0.
  2. Assemble. f′(x)=12x3−4x+5f'(x) = 12x^3 - 4x + 5.
  3. Count. Four terms in, three out, and the missing one is the constant.
  4. Check a value. At x=1x = 1, f′=13f' = 13, and the graph of ff rises steeply there, which matches.

Exam Variation (Context / Parameter Shift): A quantity is modeled by Q(t)=t44−3t2+8Q(t) = \frac{t^4}{4} - 3t^2 + 8 for t≥0t \ge 0. Find the rate of change at t=2t = 2, with units of QQ per unit of tt, and determine whether QQ is increasing or decreasing there.

Solution: The coefficient 14\frac{1}{4} passes through: Q′(t)=t3−6tQ'(t) = t^3 - 6t. At t=2t = 2 this gives 8−12=−48 - 12 = -4 units of QQ per unit of tt. The negative sign means QQ is decreasing at that instant, and the size says it is falling four units per unit of time.

Common Traps & Failure Signals:

  1. The surviving constant (Signal: an extra term in the derivative): the constant rule was skipped.
  2. The swallowed coefficient (Signal: a derivative exactly the size of the power rule's output without the multiplier): the constant multiple was divided out instead of carried.
  3. The sign drift (Signal: a plus where the original had a minus): the subtraction was not attached to the following block.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

  • Constant Kept: You carried an additive constant into the derivative.
  • Coefficient Lost →\rightarrow Say the multiplier out loud before differentiating the block.
  • Rules Blended: You applied the sum rule to a product or a quotient, which happens when the three rules are memorised as slogans rather than as statements about separate blocks.

Past 10-Year AP Exam Appearance

Testing Frequency
Rare in FRQs
The rules are the grammar of every derivative rather than an examined object, and released BC parts usually supply the derivative they intend you to use
10-Year Total Count
4 of 9 Exams · 4 FRQ Parts
Counted where a constant, a constant multiple, or a sum survives all the way into the reported derivative
FRQ Years Appeared
2018 (Q2a), 2019 (Q1d), 2023 (Q1d), 2026 (Q4A)

Practice Examples

MCQ 1

A Constant Term

If f(x)=4x3−6x+11f(x) = 4x^3 - 6x + 11, what is f′(x)f'(x)?

MCQ 2

A Coefficient

If g(x)=x52g(x) = \frac{x^5}{2}, what is g′(x)g'(x)?

MCQ 3

What the Rules Cannot Do

Which statement is true for differentiable functions ff and gg?

FRQ 1 • A Quantity Model

A quantity is modeled by Q(t)=t44−3t2+8Q(t) = \frac{t^4}{4} - 3t^2 + 8 for 0≤t≤40 \le t \le 4, where QQ is measured in liters and tt in minutes.

(a) Find Q′(t)Q'(t).

(b) Find the rate of change of QQ at t=2t = 2 minutes, with units, and state whether QQ is increasing or decreasing there.

(c) Find every time in 0<t<40 < t < 4 at which Q′(t)=0Q'(t) = 0.

(d) Explain why the constant 88 does not appear in Q′(t)Q'(t), and state what the value of QQ at t=0t = 0 contributes to the rate of change.

Cheatsheet

Basic Rules

StatementExample
Constantddx[c]=0\frac{d}{dx}[c] = 0ddx[9]=0\frac{d}{dx}[9] = 0
Constant multipleddx[cf]=cf′\frac{d}{dx}[cf] = cf'ddx[5x3]=15x2\frac{d}{dx}[5x^3] = 15x^2
Sumddx[f+g]=f′+g′\frac{d}{dx}[f+g] = f'+g'ddx[x2+x]=2x+1\frac{d}{dx}[x^2 + x] = 2x + 1
Differenceddx[f−g]=f′−g′\frac{d}{dx}[f-g] = f'-g'ddx[x2−x]=2x−1\frac{d}{dx}[x^2 - x] = 2x - 1
False product(fg)′≠f′g′(fg)' \neq f'g'x⋅xx \cdot x gives 2x2x, not 11
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