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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The constant rule, constant multiple rule, sum rule, and difference rule for differentiation, derivative notation conventions, algebraic differentiation of polynomial combinations, and application to linear rate of change problems.

You should already know: The limit definition of the derivative, basic limit evaluation rules, algebraic manipulation of polynomial functions.

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 BC 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 collection of rules are the first algebraic differentiation tools you will learn, replacing the tedious limit definition of the derivative for routine derivative calculations. Per the AP Calculus BC Course and Exam Description (CED), this content falls within Unit 2: Differentiation: Definition and Fundamental Properties, which accounts for 10–12% of the total AP exam score. These rules are tested in both multiple-choice (MCQ) and free-response (FRQ) sections, almost always as a foundational step in larger problems (e.g., finding tangent lines, critical points, or related rates) rather than as a standalone question. The core idea is that differentiation is a linear operation, meaning it follows linear rules for constants, sums, differences, and constant multiples, so we can differentiate term-by-term instead of working from first principles every time. Synonyms sometimes used for this collection are the “linearity properties of differentiation.” Even for more complex functions like products, quotients, or composites, you will rely on these basic rules to simplify your work after applying other differentiation techniques.

2. The Constant Rule

The constant rule is the simplest differentiation rule, and it follows directly from the geometric and intuitive meaning of the derivative. The derivative of a function at a point is the slope of the tangent line at that point, which measures the instantaneous rate of change of the function. A constant function $f(x)=c$ (where $c$ is any fixed real number) does not change as $x$ changes, so its rate of change is always zero. Geometrically, the graph of $f(x)=c$ is a horizontal line, which has a constant slope of zero. We can confirm this with the limit definition of the derivative: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{c - c}{h} = \lim_{h \to 0} 0 = 0 $$ In Leibniz notation, the constant rule is written as $\frac{d}{dx}[c]=0$.

Worked Example

Find $\frac{d}{dx}\left[12\pi -e^2\right]$.

1. Confirm that the expression being differentiated is a constant with respect to $x$. $12\pi \approx 37.7$ and $e^2\approx 7.39$ are both fixed real numbers, so their difference $12\pi -e^2\approx 30.31$ is also a constant independent of $x$.
2. Recall the constant rule for differentiation: $\frac{d}{dx}[c]=0$ for any constant $c\in \mathbb{R}$.
3. Substitute $c=12\pi -e^2$ into the rule.
4. The final result is $\frac{d}{dx}\left[12\pi -e^2\right]=0$.

Exam tip: If you see a constant written as a non-alphabetic symbol like $\pi$, $e$, or $g$, do not treat it as a variable unless the problem explicitly says it is a function of $x$. All fixed constants have derivative 0.

3. The Constant Multiple Rule

The constant multiple rule describes what happens when you scale a differentiable function $f(x)$ by a fixed constant $c$. The rule states that the derivative of a constant multiple of a function is equal to the constant multiple of the derivative of the function. Formally, for any constant $c$ and differentiable function $f(x)$: $$ \frac{d}{dx}\left[c \cdot f(x)\right] = c \cdot \frac{d}{dx}\left[f(x)\right] $$ We can derive this directly from the limit definition of the derivative by factoring out the constant from the limit, which is allowed by basic limit properties: $$ \lim_{h \to 0} \frac{c f(x+h) - c f(x)}{h} = c \cdot \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = c f'(x) $$ Intuitively, if you scale a function vertically by a factor of $c$, every tangent line to the function also scales vertically by $c$, so the slope (derivative) of the tangent line scales by the same factor.

Worked Example

Given that $g(x)$ is a differentiable function with $\frac{d}{dx}[g(x)]=3x^2+2x$, find $\frac{d}{dx}\left[-4g(x)\right]$.

1. Identify that we have a constant multiple of a differentiable function, with constant $c=-4$ and inner function $g(x)$.
2. Apply the constant multiple rule: $\frac{d}{dx}\left[cg(x)\right]=c\cdot \frac{d}{dx}\left[g(x)\right]$.
3. Substitute $c=-4$ and the given derivative of $g(x)$: $\frac{d}{dx}\left[-4g(x)\right]=-4\left(3x^2+2x\right)$.
4. Distribute the constant to simplify the final result: $-12x^2-8x$.

Exam tip: Always carry the negative sign through when your constant multiple is negative. A common mistake is dropping the negative when simplifying, which is an easy point deduction on FRQs.

4. The Sum and Difference Rules

The sum and difference rules extend linearity to combinations of two or more differentiable functions. The sum rule states that the derivative of a sum of two differentiable functions is the sum of their individual derivatives. The difference rule states that the derivative of a difference of two differentiable functions is the difference of their individual derivatives. Formally: $$ \frac{d}{dx}\left[f(x) + g(x)\right] = \frac{d}{dx}\left[f(x)\right] + \frac{d}{dx}\left[g(x)\right] \quad \text{(Sum Rule)} $$ $$ \frac{d}{dx}\left[f(x) - g(x)\right] = \frac{d}{dx}\left[f(x)\right] - \frac{d}{dx}\left[g(x)\right] \quad \text{(Difference Rule)} $$ These rules extend to any finite number of terms, which means we can differentiate any polynomial term-by-term. Intuitively, the total rate of change of a combination of quantities is the sum (or difference) of the individual rates of change. For example, if your total distance traveled is the sum of distance driven and distance walked, your total speed is the sum of your driving speed and walking speed.

Worked Example

Find the derivative of $f(x)=5x^3-3x^2+7x-12$, using the fact that $\frac{d}{dx}[x^n]=n{x}^{n-1}$ for any constant $n$.

1. Split the function into individual terms using the sum and difference rules: $f^{\prime }(x)=\frac{d}{dx}[5x^3]-\frac{d}{dx}[3x^2]+\frac{d}{dx}[7x]-\frac{d}{dx}[12]$.
2. Apply the constant multiple rule to each non-constant term: $f^{\prime }(x)=5\frac{d}{dx}[x^3]-3\frac{d}{dx}[x^2]+7\frac{d}{dx}[x]-\frac{d}{dx}[12]$.
3. Apply the power rule to each power of $x$ and the constant rule to the final term: $f^{\prime }(x)=5(3x^2)-3(2x)+7(1)-0$.
4. Simplify by multiplying constants to get the final derivative: $f^{\prime }(x)=15x^2-6x+7$.

Exam tip: When differentiating polynomials, always work term by term to avoid missing any terms or incorrectly combining constants. Even if you can do it in your head, writing down term-by-term differentiation reduces errors on exam day.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Differentiating the constant $\pi$ as $\frac{d}{dx}[\pi ]=1$, treating it like the variable $x$. Why: $\pi$ is written with a Greek letter that looks like a variable, so students confuse it with the independent variable $x$. Correct move: Always confirm if a symbol represents a fixed constant or a variable before differentiating; all fixed constants ($\pi$, $e$, $g$, etc.) have derivative 0.
• Wrong move: For $\frac{d}{dx}[-f(x)]$, writing $f^{\prime }(x)$ instead of $-f^{\prime }(x)$, dropping the negative sign when applying the constant multiple rule. Why: Students often focus on differentiating the function and forget to carry the negative sign from the original constant multiple. Correct move: Write the constant multiple (including negative sign) outside the derivative of the function before you differentiate, so the sign is not lost.
• Wrong move: For $\frac{d}{dx}[f(x)+c]$, writing $f^{\prime }(x)+1$ instead of $f^{\prime }(x)$, differentiating the constant term $c$ as 1. Why: Students incorrectly extend the rule $\frac{d}{dx}[x]=1$ to all constant terms, confusing the constant $c$ with the variable $x$. Correct move: Remind yourself that any term without the independent variable $x$ is a constant, so its derivative is 0, not 1.
• Wrong move: When applying the difference rule to $f(x)=3x^2-2x+5$, writing $f^{\prime }(x)=6x^2-2x$ instead of $6x-2$, because the negative sign is only applied to the coefficient and not the power rule step. Why: Students misapply signs when multiple terms are subtracted, leading to incorrect exponents or coefficients. Correct move: Assign the correct sign to each term before differentiating, so $-2x$ is a single term with constant multiple $-2$, which is carried through the entire differentiation step.
• Wrong move: Extending the constant multiple rule to $\frac{d}{dx}[f(cx)]=c{f}^{\prime }(x)$ when $c$ is inside the function argument, not a multiple of the whole function. Why: Students confuse the constant multiple rule (scaling the entire function vertically) with the constant coefficient inside a composite function, which requires the chain rule. Correct move: Only apply the constant multiple rule when the constant is multiplied by the entire function; if the constant multiplies $x$ inside the function, you must use the chain rule.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

If $f(x)=-2(3x^2-5x+7)$, what is $f^{\prime }(x)$? A) $-6x^2+10x-14$ B) $-12x+10$ C) $-12x-10$ D) $6x-5$

Worked Solution: We first recognize that $f(x)$ is a constant (-2) multiplied by a polynomial, so we apply the constant multiple rule first. By the constant multiple rule: $f^{\prime }(x)=-2\cdot \frac{d}{dx}[3x^2-5x+7]$. Next, apply the sum and difference rules to split the derivative into term-by-term calculations: $f^{\prime }(x)=-2\left(\frac{d}{dx}[3x^2]-\frac{d}{dx}[5x]+\frac{d}{dx}[7]\right)$. Applying the power rule and constant rule gives $\frac{d}{dx}[3x^2]=6x$, $\frac{d}{dx}[5x]=5$, $\frac{d}{dx}[7]=0$. Substituting back and simplifying: $f^{\prime }(x)=-2(6x-5)=-12x+10$. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=a{x}^3-b{x}^2+4$, where $a$ and $b$ are non-zero constants. (a) Use the constant, constant multiple, sum, and difference rules to find $f^{\prime }(x)$ in terms of $a$, $b$, and $x$. (b) Given that $f^{\prime }(1)=3$ and $f^{\prime }(2)=16$, find the values of $a$ and $b$. (c) Find the slope of the line tangent to $f(x)$ at $x=0$.

Worked Solution: (a) Split the function into terms using the sum and difference rules: $$f^{\prime }(x)=\frac{d}{dx}[a{x}^3]-\frac{d}{dx}[b{x}^2]+\frac{d}{dx}[4]$$ Apply the constant multiple rule to the first two terms and the constant rule to the last term: $$f^{\prime }(x)=a(3x^2)-b(2x)+0=3a{x}^2-2bx$$

(b) Substitute the given derivative values to create a system of equations: For $x=1$: $3a(1{)}^2-2b(1)=3a-2b=3$ For $x=2$: $3a(2{)}^2-2b(2)=12a-4b=16⟹3a-b=4$ Subtract the first equation from the second: $(3a-b)-(3a-2b)=4-3⟹b=1$. Substitute $b=1$ back to get $3a-2(1)=3⟹a=\frac{5}{3}$. Final result: $a=\frac{5}{3}$, $b=1$.

(c) The slope of the tangent line at $x=0$ equals $f^{\prime }(0)$. Substitute $x=0$ into $f^{\prime }(x)$: $f^{\prime }(0)=3a(0{)}^2-2b(0)=0$. The slope of the tangent line at $x=0$ is $0$.

Question 3 (Application / Real-World Style)

The total monthly profit for a small business selling custom mugs is given by $P(n)=-0.002n^2+12n-1500$, where $P(n)$ is profit in dollars, and $n$ is the number of mugs sold per month. Use the differentiation rules from this chapter to find the marginal profit function $P^{\prime }(n)$, and calculate the marginal profit when 2000 mugs are sold. Explain what your result means in context.

Worked Solution: Differentiate $P(n)$ term by term using the sum, difference, and constant multiple rules: $$ \begin{align*} P'(n) &= \frac{d}{dn}[-0.002n^2] + \frac{d}{dn}[12n] - \frac{d}{dn}[1500] \ &= -0.002 \cdot \frac{d}{dn}[n^2] + 12 \cdot \frac{d}{dn}[n] - 0 \ &= -0.002(2n) + 12(1) = -0.004n + 12 \end{align*} $$ Substitute $n=2000$ to find the marginal profit: $P^{\prime }(2000)=-0.004(2000)+12=-8+12=4$. In context, this means that when the business has already sold 2000 mugs in a month, selling one additional mug will increase their total monthly profit by approximately 4 dollars.

7. Quick Reference Cheatsheet

8. What's Next

The constant, sum, difference, and constant multiple rules are the foundational linearity properties that make all further differentiation techniques possible. Almost every derivative calculation you do for the rest of the course will rely on these rules to simplify your final result, whether you are working with products, quotients, composites, or transcendental functions. Next, you will apply these basic rules alongside the power rule to differentiate any polynomial quickly, before moving on to more advanced rules for products, quotients, and composite functions. Without mastering these basic rules, you will struggle with sign errors, constant errors, and missing terms in all subsequent differentiation work, which leads to easy point deductions on exam questions. These rules also underpin integration rules later in the course, since integration reverses differentiation and shares the same linearity properties.

Power Rule for Differentiation Product and Quotient Rules The Chain Rule Linearity of Integration