Constant, sum, difference, and constant multiple rules

AP Calculus BC· AP Calculus BC CED — Differentiation: Definition and Fundamental Properties· 14 min read

1. The Constant Rule★☆☆☆☆⏱ 3 min

📘 Definition

Constant Rule

\frac{d}{dx}[c] = 0

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.

Example:

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

🔬 Derivation

Goal:

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
$0$

Result:

This confirms the constant rule is correct.

📐 Worked Example

Find $\frac{d}{d x} [12 π - e^{2}]$

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

Exam tip:

Any fixed symbol (like $π$ , $e$ , 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.

2. The Constant Multiple Rule★★☆☆☆⏱ 3 min

📘 Definition

Constant Multiple Rule

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

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.

Example:

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

🔬 Derivation

Goal:

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
$c f'(x)$

Result:

The constant multiple rule is proven.

📐 Worked Example

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

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.

3. The Sum and Difference Rules★★☆☆☆⏱ 3 min

📘 Definition

Sum and Difference Rules

\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)

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.

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

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.

4. Combined AP-Style Worked Examples★★☆☆☆⏱ 5 min

📐 Worked Example

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

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) $f^{'} (x) = 3 a x^{2} - 2 b x$ , (b) $a = \frac{5}{3}, b = 1$ , (c) slope = 0.

📐 Worked Example

Application: The total monthly profit for a business is $P (n) = - 0.002 n^{2} + 12 n - 1500$ , 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.

5. Common Pitfalls

Wrong move:

Differentiating the constant $π$ as $\frac{d}{d x} [π] = 1$ , treating it like the variable $x$ .

Why:

$π$ 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 ( $π$ , $e$ , $g$ , etc.) have derivative 0.

Wrong move:

For $\frac{d}{d x} [- f (x)]$ , writing $f^{'} (x)$ instead of $- f^{'} (x)$ , 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 $\frac{d}{d x} [f (x) + c]$ , writing $f^{'} (x) + 1$ instead of $f^{'} (x)$ , differentiating the constant term $c$ as 1.

Why:

Students incorrectly extend the rule $\frac{d}{d x} [x] = 1$ 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 $f (x) = 3 x^{2} - 2 x + 5$ , writing $f^{'} (x) = 6 x^{2} - 2 x$ instead of $6 x - 2$ , 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 $- 2 x$ is a single term with constant multiple $- 2$ , which is carried through the entire differentiation step.

Wrong move:

Extending the constant multiple rule to $\frac{d}{d x} [f (c x)] = c f^{'} (x)$ 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.

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. Quick Reference Cheatsheet

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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Foundational step in tangent problem
2022 · FRQ
Marginal profit calculation

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.