Constant, sum, difference, and constant multiple rules — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The constant rule, constant multiple rule, sum rule, and difference rule for differentiation, derivation from the limit definition, application to polynomials, and identification of common errors on AP exam questions.

You should already know: Limit definition of the derivative, basic polynomial algebra, properties of limits.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Constant, sum, difference, and constant multiple rules?

This topic is the first set of general differentiation rules taught after introducing the derivative via limits, and it forms the core of the linearity of differentiation, a fundamental property that underpins all of calculus. Per the College Board AP Calculus AB CED, Unit 2 (Differentiation: Definition and Fundamental Properties) makes up 10–12% of the total exam score, and this topic contributes to roughly a third of that unit’s testable content, appearing in both multiple-choice (MCQ) and free-response (FRQ) sections. It is most often used as a foundational step for more complex problems like tangent line equations, related rates, and optimization, though it can also be tested directly in early MCQ questions. These rules eliminate the need to use the limit definition of the derivative for every new function, drastically simplifying differentiation of linear combinations of functions, which form the basis of most common functions on the exam. Textbooks sometimes refer to this collection of rules as the "linearity of differentiation," since they prove differentiation is a linear operation.

2. The Constant Rule

The constant rule is the simplest differentiation rule, applying to functions that are constant, meaning they output the same value for every input and have no dependence on the input variable $x$. The rule is derived directly from the limit definition of the derivative. If $f(x)=c$, where $c$ is any real constant, then by the definition of the derivative: $$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{\lim }\frac{c-c}{h}=\underset{h\to 0}{\lim }0=0$$ The formal rule is: The derivative of any constant is zero. Intuitively, a constant function graphs as a horizontal line, which has a slope of zero everywhere. Since the derivative at a point is the slope of the tangent line at that point, a slope of zero everywhere means the derivative is always zero.

Worked Example

Problem: Find the derivative of $g(x)=-\frac{9\pi }{4}$ with respect to $x$.

1. Confirm that $-\frac{9\pi }{4}$ is a constant: it has no dependence on the input variable $x$, so it fits the definition of a constant function.
2. Recall the constant rule formula: $\frac{d}{dx}[c]=0$ for any real constant $c$.
3. Substitute $c=-\frac{9\pi }{4}$ into the rule.
4. Final result: $\frac{d}{dx}\left[-\frac{9\pi }{4}\right]=0$.

Exam tip: On the AP exam, constants like $\pi$, $e$, and arbitrary constants labeled $k$ in problems are still constants — always apply the constant rule to them, don't mistake them for variables.

3. The Constant Multiple Rule

The constant multiple rule extends differentiation to terms that have a constant coefficient multiplied by a non-constant function. The rule states that the derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function. It is derived from the constant multiple property of limits: if $c$ is a constant and $f(x)$ is differentiable, then: $$\frac{d}{dx}\left[c\cdot f(x)\right]=\underset{h\to 0}{\lim }\frac{cf(x+h)-cf(x)}{h}=c\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}=c\cdot f^{\prime }(x)$$ Intuitively, if you vertically stretch or compress a function by a factor of $c$, every tangent line’s stretch or compresses by the same factor $c$, so the slope changes by exactly $c$. This rule lets us pull constant coefficients out of derivatives, which is essential for term-by-term differentiation of polynomials.

Worked Example

Problem: Given that $\frac{d}{dx}[x^3]=3x^2$, use the constant multiple rule to find the derivative of $y=-5x^3$.

1. Rewrite $y$ to match the constant multiple rule form: $y=-5\cdot f(x)$, where $c=-5$ and $f(x)=x^3$.
2. Apply the constant multiple rule: $\frac{dy}{dx}=c\cdot f^{\prime }(x)=-5\cdot \frac{d}{dx}[x^3]$.
3. Substitute the known derivative $\frac{d}{dx}[x^3]=3x^2$.
4. Simplify: $\frac{dy}{dx}=-5(3x^2)=-15x^2$.

Exam tip: Never forget that negative constants are still constants — keep the negative sign with the constant when you pull it out, don't accidentally drop it during simplification.

4. The Sum and Difference Rules

The sum and difference rules let us differentiate combinations of two or more differentiable functions term-by-term, which is required to differentiate any polynomial. Formally, if $f(x)$ and $g(x)$ are both differentiable at $x$, then: $$\frac{d}{dx}\left[f(x)+g(x)\right]=\frac{d}{dx}\left[f(x)\right]+\frac{d}{dx}\left[g(x)\right]$$ $$\frac{d}{dx}\left[f(x)-g(x)\right]=\frac{d}{dx}\left[f(x)\right]-\frac{d}{dx}\left[g(x)\right]$$ These rules follow directly from the limit property that the limit of a sum is the sum of the limits. The difference rule is actually a combination of the constant multiple rule (with $c=-1$) and the sum rule: $f(x)-g(x)=f(x)+(-1)g(x)$, so its derivative is $f^{\prime }(x)+(-1)g^{\prime }(x)=f^{\prime }(x)-g^{\prime }(x)$. Together with the constant rule, constant multiple rule, and the power rule, these rules let us differentiate any polynomial without using the limit definition.

Worked Example

Problem: Differentiate $f(x)=5x^4-3x^2+2x-9$ with respect to $x$, given that $\frac{d}{dx}[x^n]=n{x}^{n-1}$.

1. Split the function into individual terms using the sum and difference rules: $$\frac{d}{dx}[5x^4-3x^2+2x-9]=\frac{d}{dx}[5x^4]-\frac{d}{dx}[3x^2]+\frac{d}{dx}[2x]-\frac{d}{dx}[9]$$
2. Apply the constant multiple rule to each variable term, and the constant rule to the final term: $$=5\cdot \frac{d}{dx}[x^4]-3\cdot \frac{d}{dx}[x^2]+2\cdot \frac{d}{dx}[x]-0$$
3. Substitute the derivative for each power of $x$: $$=5(4x^3)-3(2x)+2(1)-0$$
4. Simplify: $f^{\prime }(x)=20x^3-6x+2$.

Exam tip: Always check that you have differentiated every term in the function, including the final constant term — it is easy to forget it and accidentally carry it over into the final derivative.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Differentiating $y=7$ to get $y^{\prime }=7$. Why: Students confuse the constant value itself with its derivative, forgetting the constant rule always outputs zero. Correct move: Mark any term without an input variable as a constant, and write 0 for its derivative immediately before moving on.
• Wrong move: Differentiating $y=-4x^2$ to get $y^{\prime }=-8x^2$. Why: Students only apply the derivative to the variable part and leave the original power of $x$ unchanged, forgetting to multiply the constant coefficient by the derivative of the variable part. Correct move: After finding the derivative of the variable part, always multiply it by the full constant coefficient, including its sign.
• Wrong move: Differentiating $f(x)=\pi x$ to get $f^{\prime }(x)=x+\pi$. Why: Students confuse the constant $\pi$ with a variable and unnecessarily apply the product rule. Correct move: All Greek letters and arbitrary labeled constants ($k,a,b$) are constants, so apply the constant multiple rule to pull them out.
• Wrong move: Writing $\frac{d}{dx}[3x+4]=3+4=7$. Why: Students forget to apply the constant rule to the constant term and incorrectly carry the constant into the derivative. Correct move: Cross out constant terms after replacing their derivative with 0 to avoid accidental carryover.
• Wrong move: Differentiating $f(x)=x^3-2x^2+x-4$ to get $f^{\prime }(x)=3x^2-4x+1-4=3x^2-4x-3$. Why: Students fail to replace the final constant with zero, leaving it in the final derivative. Correct move: Count the number of terms in the original function and confirm you have one fewer non-zero term in the derivative (for polynomials with a constant term).

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

If $f(x)=\frac{1}{3}{x}^6-9x+5e$, what is $f^{\prime }(x)$? A) $2x^5-9+5e$ B) $2x^5-9$ C) $2x^6-9+5$ D) $\frac{1}{3}\cdot 6x^5-9$

Worked Solution: Split $f(x)$ into terms using the sum and difference rules to get $f^{\prime }(x)=\frac{d}{dx}\left[\frac{1}{3}{x}^6\right]-\frac{d}{dx}[9x]+\frac{d}{dx}[5e]$. Apply the constant multiple rule to the first two terms, and the constant rule to $5e$, which is a constant. Using the power rule $\frac{d}{dx}[x^6]=6x^5$ and $\frac{d}{dx}[x]=1$, we get $\frac{1}{3}(6x^5)-9(1)+0=2x^5-9$. The distractors A incorrectly leaves $5e$ undifferentiated, C incorrectly differentiates $e$ as 1, and D is unsimplified. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=8x^3-12x^2+4x-10$. (a) Use the constant, sum, difference, and constant multiple rules to find $f^{\prime }(x)$. You may use the power rule $\frac{d}{dx}[x^n]=n{x}^{n-1}$. (b) Find the slope of the tangent line to $f(x)$ at $x=2$. (c) Write the equation of the tangent line to $f(x)$ at $x=2$.

Worked Solution: (a) Split into terms by the sum/difference rules: $$f^{\prime }(x)=\frac{d}{dx}[8x^3]-\frac{d}{dx}[12x^2]+\frac{d}{dx}[4x]-\frac{d}{dx}[10]$$ Apply the constant multiple and power rules to variable terms, and the constant rule to the constant term: $$f^{\prime }(x)=8(3x^2)-12(2x)+4(1)-0=24x^2-24x+4$$

(b) The slope of the tangent at $x=2$ equals $f^{\prime }(2)$: $$f^{\prime }(2)=24(2{)}^2-24(2)+4=24(4)-48+4=96-48+4=52$$ The slope is $52$.

(c) Find the point on $f(x)$ at $x=2$: $f(2)=8(8)-12(4)+4(2)-10=64-48+8-10=14$, so the point is $(2,14)$. Use point-slope form $y-y_1=m(x-x_1)$: $$y-14=52(x-2)⟹y=52x-104+14=52x-90$$ The tangent line equation is $y=52x-90$.

Question 3 (Application / Real-World Style)

The monthly profit from selling $x$ custom bicycles is given by $P(x)=-150x^2+6000x-350$, where $P(x)$ is profit in US dollars and $x$ is the number of bicycles sold. The marginal profit at a sales level of $x$ bicycles is defined as the derivative of the profit function at $x$, and it approximates the change in profit from selling one additional bicycle. Find the marginal profit when 15 bicycles are sold, including units.

Worked Solution: First find the marginal profit function $MP(x)=P^{\prime }(x)$ using our differentiation rules: $$MP(x)=\frac{d}{dx}[-150x^2]+\frac{d}{dx}[6000x]-\frac{d}{dx}[350]=-150(2x)+6000(1)-0=-300x+6000$$ Substitute $x=15$ to find marginal profit at 15 bicycles: $$MP(15)=-300(15)+6000=-4500+6000=1500$$ In context, when 15 bicycles have been sold, the profit will increase by approximately $1500 from selling one additional bicycle.

7. Quick Reference Cheatsheet

8. What's Next

These four rules are the foundation for all differentiation you will do for the rest of AP Calculus AB. Every more advanced rule, from the product rule to the chain rule, relies on the linearity of differentiation that we formalized here. Without mastering these rules, you will make consistent sign errors, constant errors, and term-by-term differentiation errors on nearly every differentiation problem for the rest of the course. Immediately next, you will apply these rules along with the power rule to differentiate any polynomial, and later use these rules as a routine step in more complex problems like implicit differentiation and related rates. This topic is also the basis for all linearity of integration rules later in the course, since integration inherits this property directly from differentiation.

Power rule for differentiation Product and quotient rules Tangent line approximation Derivatives of trigonometric functions