# Constant, sum, difference, and constant multiple rules

> AP Calculus BC · Unit 2: Differentiation: Definition and Fundamental Properties
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u2-constant-sum-difference-and-constant/

This module covers the four core linear differentiation rules: constant, constant multiple, sum, and difference. You will learn to apply these rules to replace tedious limit-based differentiation for routine calculations, a foundational skill for all AP Calculus BC problems.

**Prerequisites:** [Limit definition of the derivative](https://www.owlsprep.com/study/ap-calculus-bc-u2-derivative-definition/); Basic limit evaluation rules; Algebraic manipulation of polynomial functions

## Learning objectives

- State the constant, constant multiple, sum, and difference differentiation rules
- Apply the rules to differentiate polynomial combinations of functions
- Identify and avoid common sign and constant errors in differentiation
- Use these rules to solve applied rate of change problems

## The Constant Rule

**Constant Rule** — For any fixed real constant $c$, the derivative of the constant function $f(x) = c$ is zero. This follows from the fact that constant functions have zero rate of change.

*Notation:* \frac{d}{dx}[c] = 0

*Example:* The graph of $f(x) = c$ is a horizontal line with slope 0, confirming the rule.

**Derivation:** Prove the constant rule from the limit definition of the derivative

*Starting from:* The limit definition: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

1. Substitute $f(x) = c$ into the definition:
2. $$\lim_{h \to 0} \frac{c - c}{h}$$
3. Simplify the numerator:
4. $$\lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0$$
5. Evaluate the limit of a constant, which equals the constant:
6. $$0$$

*Conclusion:* This confirms the constant rule is correct.

**Worked example:** Find $\frac{d}{dx}\left[12\pi - e^2\right]$

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

> **Exam tip:** Any fixed symbol (like $\pi$, $e$, or $g$) is a constant, not a variable, unless the problem explicitly states it is a function of $x$. All fixed constants have derivative 0.

## The Constant Multiple Rule

**Constant Multiple Rule** — The derivative of a constant multiple of a differentiable function equals the constant multiple of the function's derivative. This follows from the constant multiple rule for limits.

*Notation:* \frac{d}{dx}\left[c \cdot f(x)\right] = c \cdot \frac{d}{dx}\left[f(x)\right]

*Example:* Scaling a function vertically scales all tangent slopes by the same constant factor.

**Derivation:** Prove the constant multiple rule from the limit definition of the derivative

*Starting from:* Limit definition of derivative for $c f(x)$

1. Write the derivative using the definition:
2. $$\lim_{h \to 0} \frac{c f(x+h) - c f(x)}{h}$$
3. Factor the constant $c$ out of the numerator and limit, per limit properties:
4. $$c \cdot \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
5. Recognize the limit as the derivative of $f(x)$:
6. $$c f'(x)$$

*Conclusion:* The constant multiple rule is proven.

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

1. Identify $c = -4$ (the constant multiple) and inner function $g(x)$.
2. Apply the constant multiple rule:
3. $$\frac{d}{dx}\left[-4 g(x)\right] = -4 \cdot \frac{d}{dx}\left[g(x)\right]$$
4. Substitute the given derivative of $g(x)$ and simplify:
5. $$-4(3x^2 + 2x) = -12x^2 - 8x$$

> **Exam tip:** Always carry negative signs through when applying the constant multiple rule to negative constants. Dropping the negative is a common source of exam point loss.

## The Sum and Difference Rules

**Sum and Difference Rules** — The derivative of a sum (or difference) of two differentiable functions equals the sum (or difference) of their individual derivatives. These rules extend to any finite number of terms, enabling term-by-term differentiation.

*Notation:* \frac{d}{dx}\left[f(x) + g(x)\right] = f'(x) + g'(x) \\ \frac{d}{dx}\left[f(x) - g(x)\right] = f'(x) - g'(x)

*Example:* If total speed is the sum of two individual speeds, the total rate of change is the sum of the individual rates.

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

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

> **Exam tip:** When differentiating polynomials, work term by term explicitly to avoid missing terms or making constant errors. Even if you can do it in your head, writing it down reduces mistakes on exam day.

## Combined AP-Style Worked Examples

**Worked example:** Multiple Choice: If $f(x) = -2(3x^2 - 5x + 7)$, what is $f'(x)$? Options: A) $-6x^2 + 10x - 14$, B) $-12x + 10$, C) $-12x - 10$, D) $6x - 5$

1. Recognize $f(x)$ is a constant (-2) multiplied by a polynomial. Apply the constant multiple rule first:
2. $$f'(x) = -2 \cdot \frac{d}{dx}[3x^2 - 5x + 7]$$
3. Apply sum and difference rules to split into term-by-term derivatives:
4. $$f'(x) = -2\left(\frac{d}{dx}[3x^2] - \frac{d}{dx}[5x] + \frac{d}{dx}[7]\right)$$
5. Apply power and constant rules to each term:
6. $$\frac{d}{dx}[3x^2] = 6x, \quad \frac{d}{dx}[5x] = 5, \quad \frac{d}{dx}[7] = 0$$
7. Substitute and simplify:
8. $$f'(x) = -2(6x - 5) = -12x + 10$$
9. The correct answer is B.

**Worked example:** Free Response: Let $f(x) = a x^3 - b x^2 + 4$, where $a$ and $b$ are non-zero constants. (a) Find $f'(x)$ using the rules from this module. (b) Given $f'(1) = 3$ and $f'(2) = 16$, find $a$ and $b$. (c) Find the slope of the tangent to $f(x)$ at $x=0$.

1. Part (a): Split into terms and apply the rules:
2. $$f'(x) = \frac{d}{dx}[a x^3] - \frac{d}{dx}[b x^2] + \frac{d}{dx}[4] = 3a x^2 - 2b x$$
3. Part (b): Substitute the given conditions to get a system of equations:
4. $$3a(1)^2 - 2b(1) = 3 \implies 3a - 2b = 3 \\ 3a(2)^2 - 2b(2) = 16 \implies 12a - 4b = 16 \implies 3a - b = 4$$
5. Solve the system: subtract the first equation from the second to get $b = 1$, then substitute back to find $a = \frac{5}{3}$.
6. Part (c): The slope of the tangent at $x=0$ equals $f'(0)$:
7. $$f'(0) = 3a(0)^2 - 2b(0) = 0$$
8. Final results: (a) $f'(x) = 3a x^2 - 2b x$, (b) $a = \frac{5}{3}, b=1$, (c) slope = 0.

**Worked example:** Application: The total monthly profit for a business is $P(n) = -0.002n^2 + 12n - 1500$, where $P(n)$ is profit in dollars, and $n$ is number of mugs sold. Find the marginal profit function $P'(n)$, calculate marginal profit at $n=2000$, and interpret the result.

1. Differentiate term by term using the rules:
2. $$P'(n) = \frac{d}{dn}[-0.002n^2] + \frac{d}{dn}[12n] - \frac{d}{dn}[1500] \\ = -0.002(2n) + 12(1) - 0 = -0.004n + 12$$
3. Substitute $n=2000$:
4. $$P'(2000) = -0.004(2000) + 12 = 4$$
5. Interpretation: When 2000 mugs have already been sold, selling one additional mug increases total monthly profit by approximately \$4.

## Common pitfalls

- **Wrong:** Differentiating the constant $\pi$ as $\frac{d}{dx}[\pi] = 1$, treating it like the variable $x$.
  - Why it fails: $\pi$ is written with a Greek letter that looks like a variable, so students confuse it with the independent variable $x$.
  - Correct: 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:** For $\frac{d}{dx}[-f(x)]$, writing $f'(x)$ instead of $-f'(x)$, dropping the negative sign when applying the constant multiple rule.
  - Why it fails: Students often focus on differentiating the function and forget to carry the negative sign from the original constant multiple.
  - Correct: Write the constant multiple (including negative sign) outside the derivative of the function before you differentiate, so the sign is not lost.
- **Wrong:** For $\frac{d}{dx}[f(x) + c]$, writing $f'(x) + 1$ instead of $f'(x)$, differentiating the constant term $c$ as 1.
  - Why it fails: Students incorrectly extend the rule $\frac{d}{dx}[x] = 1$ to all constant terms, confusing the constant $c$ with the variable $x$.
  - Correct: Remind yourself that any term without the independent variable $x$ is a constant, so its derivative is 0, not 1.
- **Wrong:** When applying the difference rule to $f(x) = 3x^2 - 2x + 5$, writing $f'(x) = 6x^2 - 2x$ instead of $6x - 2$, because the negative sign is only applied to the coefficient and not the power rule step.
  - Why it fails: Students misapply signs when multiple terms are subtracted, leading to incorrect exponents or coefficients.
  - Correct: 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:** Extending the constant multiple rule to $\frac{d}{dx}[f(cx)] = c f'(x)$ when $c$ is inside the function argument, not a multiple of the whole function.
  - Why it fails: Students confuse the constant multiple rule (scaling the entire function vertically) with the constant coefficient inside a composite function.
  - Correct: 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.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Constant Rule | $\frac{d}{dx}[c] = 0$ | Any fixed real constant; applies to $\pi$, $e$, and all fixed coefficients |
| Constant Multiple Rule | $\frac{d}{dx}\left[c \cdot f(x)\right] = c \cdot \frac{d}{dx}\left[f(x)\right]$ | Only applies when constant multiplies the entire function; use chain rule for constants inside arguments |
| Sum Rule | $\frac{d}{dx}\left[f(x) + g(x)\right] = \frac{d}{dx}\left[f(x)\right] + \frac{d}{dx}\left[g(x)\right]$ | Extends to any finite number of summed terms; works for all differentiable functions |
| Difference Rule | $\frac{d}{dx}\left[f(x) - g(x)\right] = \frac{d}{dx}\left[f(x)\right] - \frac{d}{dx}\left[g(x)\right]$ | Equivalent to sum rule with $c=-1$ for $g(x)$; preserves term signs |
| Combined Linearity Rule | $\frac{d}{dx}\left[a f(x) + b g(x)\right] = a f'(x) + b g'(x)$ | Combines all four rules into one formula for linear combinations of functions |
| Term-by-Term Polynomial Differentiation | $\frac{d}{dx}\left[\sum_{k=0}^n c_k x^k\right] = \sum_{k=1}^n k c_k x^{k-1}$ | Constant term $c_0$ drops out because its derivative is 0 |

## What's next

These four rules are the foundational linearity properties that make all further differentiation techniques possible in AP Calculus BC. Almost every derivative calculation you will complete for the rest of the course relies on these rules to simplify your final result, whether you are working with products, quotients, composite functions, or transcendental functions. Mastering these rules now eliminates easy point errors from sign mistakes or incorrect constant handling on all future differentiation problems. These linearity properties also extend directly to integration later in the course, since integration reverses differentiation and shares the same core properties. Building a solid understanding of these basic rules will set you up for success for all more advanced topics in calculus.

- [Power Rule for Differentiation](https://www.owlsprep.com/study/ap-calculus-bc-u2-power-rule/)
- [The Chain Rule](https://www.owlsprep.com/study/ap-calculus-bc-u3-chain-rule/)
- [Derivatives of cos x, sin x, e^x, ln(x)](https://www.owlsprep.com/study/ap-calculus-bc-u2-derivatives-of-cos-sin-e/)

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