Unit 2Constant Multiple and Sum/Difference Rules

Concept 4.76Built for May 2027 AP Exam

Constant Multiple and Sum/Difference Rules

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Differentiation is linear. Two rules carry that sentence: a constant multiplier survives the derivative unchanged, $\frac{d}{dx}\bigl[cf(x)\bigr] = cf'(x)$ , and a sum or difference differentiates one term at a time, $\frac{d}{dx}\bigl[f(x) \pm g(x)\bigr] = f'(x) \pm g'(x)$ . Together with the power rule they cover every polynomial, which is why this is the first topic where long expressions stop being intimidating. Once the split is made, each piece becomes a one-line problem, and the answer is those lines rejoined in the original order. The rules say nothing about products, quotients, or compositions. Those need their own machinery, and applying the linear rules across a multiplication sign is the single most expensive mistake in Unit 2.

On the AP Exam

This is infrastructure. Nothing is ever asked about the rules themselves; they are the reason a seven-term polynomial takes twenty seconds instead of a page. The grader sees only the derivative that results, so the rules earn credit silently and lose it silently, usually when a minus sign is dropped between terms.

What you’ll see: A polynomial written out in full, a function with a fractional coefficient, or a derivative that must be read off a differential equation and then differentiated again.
What you’ll do: Split the expression at each plus or minus, keep every coefficient in place, and differentiate the pieces.
Memory hook: Constants ride along, sums break apart. Neither operation touches the structure of the expression.
Use it on the exam: For MCQs: Check the number of terms in the answer against the number in the question. A four-term polynomial that produces a two-term derivative has lost something. For FRQs: When a differential equation such as $\frac{dH}{dt} = \frac{1}{3}(27 - H)$ must be differentiated again, the constant multiple rule is the whole step: $\frac{d^2H}{dt^2} = -\frac{1}{3}\frac{dH}{dt}$ .

Concept Review

Figure 1

Linearity splits the work and keeps the coefficients. The three chips are the same rule read three ways: a polynomial, a signed sum, and a fractional multiplier.

EACH TERM KEEPS ITS OWN COEFFICIENT

The Two Rules Side by Side

Rule	Statement	What it licenses
Constant multiple	$\frac{d}{dx}\bigl[cf(x)\bigr] = cf'(x)$	A number in front passes through untouched
Sum	$\frac{d}{dx}\bigl[f(x) + g(x)\bigr] = f'(x) + g'(x)$	Split at a plus sign
Difference	$\frac{d}{dx}\bigl[f(x) - g(x)\bigr] = f'(x) - g'(x)$	Split at a minus sign, and keep the sign
Constant term	$\frac{d}{dx}[c] = 0$	A bare number disappears

The fourth line is a special case of the first, with $f(x) = 1$ . It is listed separately because students who understand $cf'$ still sometimes keep the constant term.

What Linearity Actually Says

The derivative is a limit of a quotient, and limits already respect sums and constant multiples. That is the whole proof: $\lim \bigl[cf\bigr]$ is $c \lim f$ , and $\lim\bigl[f + g\bigr]$ is $\lim f + \lim g$ , provided the individual limits exist. The rules for derivatives inherit those properties without any extra work.

$\frac{d}{dx}\Bigl[3x^4 - \frac{1}{2}x^2 + 7x - 5\Bigr] = 12x^3 - x + 7$

Four terms in, three terms out, because the constant died. Nothing was rearranged.

Where the Licence Stops

Expression	Legal move	Illegal move
$5x^3$	$5 \cdot 3x^2 = 15x^2$	$5x^2$ — the coefficient was consumed
$x^2 + x^2$	$2x + 2x = 4x$	$2x$ — combining after differentiating loses a term
$x^2 \cdot x^3$	Rewrite as $x^5$ , then $5x^4$	$2x \cdot 3x^2$ — the product rule is not this
$\frac{x^3}{x}$	Simplify to $x^2$ , then $2x$	$\frac{3x^2}{1}$ — the quotient rule is not needed after simplifying

Two rows are about boundaries rather than about this topic. The product and the quotient look like they should obey a simple rule, and they do not; that asymmetry is why Units 2.11 and 2.12 exist.

Common Misconception Check

Myth 1: "The derivative of $5f(x)$ is $f'(x)$ ." Truth: The multiplier stays. Only the function changes, so the constant is still there when the answer is written.
Myth 2: "A sum differentiates as a unit." Truth: Each term is independent. Differentiating term by term is not a shortcut; it is the rule.
Myth 3: "The rules extend to products because multiplication distributes over addition." Truth: Distribution is a property of multiplication, not of differentiation. $\frac{d}{dx}[fg]$ is not $f'g'$ , and the product rule is required.

Topic-Specific Study Method

Split, Then Scale

Borrowed idea: Which operation comes first, the scaling or the summing? The derivative is linear, so it does not care: it passes through constants and through plus signs without rearranging anything. That single fact turns a long polynomial into a column of short problems, one per term. Does the same freedom extend to multiplication? It does not, and the product rule exists precisely because it fails. Treat linearity as a licence to split, and treat every product, quotient, or composition as a boundary the licence does not cross.

Exam disguise	What you see on the page	What your brain must translate	Required execution
Seven-term polynomial	A long expression to differentiate	Seven independent short problems	Split, differentiate, rejoin in order
Fractional coefficient	$\frac{2}{3}x^6$	A constant multiple with a fraction	Multiply $\frac{2}{3}$ by $6$ to get $4x^5$
Second derivative of a model	$\frac{dH}{dt} = \frac{1}{3}(27 - H)$	A constant multiple of a difference	$\frac{d^2H}{dt^2} = -\frac{1}{3}\frac{dH}{dt}$
A difference of two rates	$h(x) = f(x) - g(x)$	A difference rule with signs kept	$h'(x) = f'(x) - g'(x)$

① Differentiate $P(x) = 0.02x^3 - 1.5x^2 + 40x + 90$ twice, and count the terms at each stage. Three, then two, then one: the polynomial is finite, and its derivatives run out.

② Take a differential equation from a released exam and differentiate both sides once. The left side becomes a second derivative and the right side exercises the constant multiple rule, which is exactly the step 2017 Q4(b) and 2023 Q3(c) reward.

③ Write the five expressions you most often meet on the exam, then mark each one with the first legal move: split, rewrite, or stop. The marking takes less time than the differentiating.

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $f(x) = \frac{3}{4}x^8 - 2x^3 + \frac{1}{2}x - 11$ .

Standard Execution:

Split. Four terms, four independent derivatives.

Scale the first. $\frac{3}{4} \cdot 8 = 6$ , so the term becomes $6x^7$ .

Scale the second. $-2 \cdot 3 = -6$ , so the term becomes $-6x^2$ .

Scale the third. $\frac{1}{2} \cdot 1 = \frac{1}{2}$ , so the term becomes $\frac{1}{2}$ .

Drop the last. The constant $-11$ contributes $0$ , leaving $f'(x) = 6x^7 - 6x^2 + \frac{1}{2}$ .

Exam Variation (Context / Parameter Shift): A quantity $H$ satisfies $\frac{dH}{dt} = \frac{1}{4}(40 - H)$ for all $t > 0$ . Determine whether the graph of $H$ is concave up or concave down at a time when $H = 20$ .

Solution: Differentiate both sides with respect to $t$ . The right side is a constant multiple of a difference, so $\frac{d^2H}{dt^2} = \frac{1}{4}\left(0 - \frac{dH}{dt}\right) = -\frac{1}{4}\frac{dH}{dt}$ . At $H = 20$ the given equation produces $\frac{dH}{dt} = \frac{1}{4}(20) = 5$ , so $\frac{d^2H}{dt^2} = -\frac{5}{4} < 0$ and the graph is concave down. Two applications of the same rule, no integration needed.

Common Traps & Failure Signals:

The swallowed coefficient (Signal: a derivative one power of the constant too small): the multiplier was treated as part of the base.
The lost minus (Signal: a subtracted term arriving positive): the difference rule was applied without carrying the sign.
The split that should not happen (Signal: two derivatives multiplied together under a product): linearity was extended past a multiplication sign.

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

Licence Overreach $\rightarrow$ Stop splitting at the first multiplication sign.
Sign Drop: You lost the minus on a subtracted term.
Coefficient Creep: You pushed the constant into the exponent instead of leaving it in front, so $5x^4$ became $x^{20}$ and every later step of the solution inherited an error that no amount of careful algebra could repair.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; The rules are applied inside second-derivative and rate-of-change parts rather than examined on their own
10-Year Total Count: 6 of 9 Exams · 6 FRQ Parts
FRQ Years Appeared: 2017 (Q4b), 2018 (Q2a), 2019 (Q1d), 2022 (Q2b), 2023 (Q3c), 2026 (Q3B)

Practice Examples

MCQ 1

Scaling

If $f(x) = \frac{4}{5}x^{10}$ , what is $f'(x)$ ?

MCQ 2

Signs

Let $g(x) = x^5 - 3x^2 + 7$ . What is $g'(x)$ ?

MCQ 3

The Boundary

Which expression may be differentiated by splitting it into separate derivatives?

FRQ 4

A Cooling Model

A cup of water is placed in a room. The temperature $T$ , in degrees Celsius, of the water at time $t$ minutes satisfies the differential equation

$\frac{dT}{dt} = -\frac{1}{5}\bigl(T - 22\bigr)$

for $t \ge 0$ , with $T(0) = 90$ .

(a) Find $\frac{d^2T}{dt^2}$ in terms of $T$ , and explain which derivative rule produces it.

(b) At the instant when $T = 42$ , determine whether the temperature is decreasing at an increasing rate or at a decreasing rate. Give a reason for your answer.

(c) Let $S(t) = 3T(t) - 5t$ . Find $S'(t)$ in terms of $T$ and $t$ .

(d) The water is replaced by a larger container whose temperature $U$ satisfies $\frac{dU}{dt} = -0.1(U - 22)$ with $U(0) = 90$ . Without solving either equation, explain which container cools faster at $t = 0$ .

Cheatsheet