Constant, sum, difference, and constant multiple rules

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

1. The Constant Rule★☆☆☆☆⏱ 3 min

The constant rule is the simplest differentiation rule, applying to constant functions that output the same value for every input and have no dependence on $x$ . It is derived directly from the limit definition of the derivative, and matches the intuitive fact that a horizontal line has a slope of 0 everywhere.

📘 Definition

Constant Rule

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

For any real constant $c$ , the derivative of the constant function $f (x) = c$ is always equal to zero.

Example:

$\frac{d}{d x} [12] = 0$ , $\frac{d}{d x} [e] = 0$

🔬 Derivation

Goal:

Derive the constant rule from the limit definition of the derivative

Starting from:

Derivative definition: $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ , for $f (x) = c$

1
Substitute $f (x + h) = c$ and $f (x) = c$ into the definition:
$\lim_{h \to 0} \frac{c - c}{h}$
2
Simplify the numerator:
$\lim_{h \to 0} \frac{0}{h} = \lim_{h \to 0} 0$
3
Evaluate the limit of a constant:
$f'(x) = 0$

Result:

The derivative of any constant is zero, confirming the constant rule.

📐 Worked Example

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

1
Confirm that $- \frac{9 π}{4}$ 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:
3
$\frac{d}{dx}[c] = 0 \text{ for any real constant } c$
4
Substitute $c = - \frac{9 π}{4}$ into the rule.
5
Final result:
6
$\frac{d}{dx}\left[-\frac{9\pi}{4}\right] = 0$

Exam tip:

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

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

The constant multiple rule extends differentiation to terms that have a constant coefficient multiplied by a non-constant function. It lets us pull constant coefficients out of derivatives, which is an essential step for term-by-term differentiation of polynomials.

📘 Definition

Constant Multiple Rule

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

If $c$ is a constant and $f (x)$ is differentiable, the derivative of $c$ times $f (x)$ equals $c$ multiplied by the derivative of $f (x)$ .

🔬 Derivation

Goal:

Derive the constant multiple rule from limit properties

Starting from:

Constant multiple property of limits and the derivative definition

1
Start with the derivative definition for $c f(x):
2
$\frac{d}{dx}\left[c f(x)\right] = \lim_{h \to 0} \frac{c f(x+h) - c f(x)}{h}$
3
Factor the constant $c$ out of the limit:
4
$c \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
5
Recognize the limit as $f^{'} (x)$ , the derivative of $f (x)$ :

Result:

Thus, $\frac{d}{d x} [c f (x)] = c f^{'} (x)$ , which proves the rule.

📐 Worked Example

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

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:
3
$\frac{dy}{dx} = c \cdot f'(x) = -5 \cdot \frac{d}{dx}[x^3]$
4
Substitute the known derivative $\frac{d}{d x} [x^{3}] = 3 x^{2}$ .
5
Simplify to get the final result:
6
$\frac{dy}{dx} = -5(3x^2) = -15x^2$

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.

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

The sum and difference rules allow us to differentiate linear combinations of functions term-by-term, which is required to differentiate any polynomial. These rules follow directly from the limit property that the limit of a sum is the sum of the limits. The difference rule can also be derived by combining the sum rule with the constant multiple rule for $c = - 1$ .

📘 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)

If $f (x)$ and $g (x)$ are both differentiable at $x$ , the derivative of a sum or difference of functions equals the sum or difference of their individual derivatives. This extends to any finite number of terms.

📐 Worked Example

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

1
Split the function into individual terms using the sum and difference rules:
2
$\frac{d}{dx}[5x^4 - 3x^2 + 2x - 9] = \frac{d}{dx}[5x^4] - \frac{d}{dx}[3x^2] + \frac{d}{dx}[2x] - \frac{d}{dx}[9]$
3
Apply the constant multiple rule to each variable term, and the constant rule to the final term:
4
$= 5 \cdot \frac{d}{dx}[x^4] - 3 \cdot \frac{d}{dx}[x^2] + 2 \cdot \frac{d}{dx}[x] - 0$
5
Substitute the power rule derivative for each power of $x$ :
6
$= 5(4x^3) - 3(2x) + 2(1) - 0$
7
Simplify to get the final derivative:
8
$f'(x) = 20x^3 - 6x + 2$

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.

4. Combined AP-Style Applications★★★☆☆⏱ 5 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

If $f (x) = \frac{1}{3} x^{6} - 9 x + 5 e$ , what is $f^{'} (x)$ ?
- A) $2 x^{5} - 9 + 5 e$
- B) $2 x^{5} - 9$
- C) $2 x^{6} - 9 + 5$
- D) $2 x^{5} - 9 e$
Reveal answer
B —
Distractor A leaves the constant $5 e$ undifferentiated, C incorrectly differentiates the constant, D incorrectly treats $e$ as a variable. The constant rule gives $\frac{d}{d x} [5 e] = 0$ , so the correct derivative is $2 x^{5} - 9$ .

📐 Worked Example

Let $f (x) = 8 x^{3} - 12 x^{2} + 4 x - 10$ . (a) Use the core rules to find $f^{'} (x)$ (you may use the power rule). (b) Find the slope of the tangent line at $x = 2$ . (c) Write the equation of the tangent line at $x = 2$ .

1
Part (a): Split into terms and apply rules:
2
$f'(x) = \frac{d}{dx}[8x^3] - \frac{d}{dx}[12x^2] + \frac{d}{dx}[4x] - \frac{d}{dx}[10]$
3
$f'(x) = 8(3x^2) - 12(2x) + 4(1) - 0 = 24x^2 - 24x + 4$
4
Part (b): Slope equals $f^{'} (2)$ . Substitute $x = 2$ :
5
$f'(2) = 24(2)^2 - 24(2) + 4 = 52$
6
Part (c): Find the point $(2, f (2))$ :
7
$f(2) = 8(8) - 12(4) + 4(2) - 10 = 14, \text{ so point } (2, 14)$
8
Use point-slope form and simplify:
9
$y - 14 = 52(x - 2) \implies y = 52x - 90$

📐 Worked Example

The monthly profit from selling $x$ custom bicycles is $P (x) = - 150 x^{2} + 6000 x - 350$ , where $P (x)$ is profit in US dollars. Marginal profit (the derivative of $P (x)$ ) approximates the change in profit from selling one additional bicycle. Find the marginal profit when 15 bicycles are sold, including units.

1
Find the marginal profit function $M P (x) = P^{'} (x)$ using the differentiation rules:
2
$MP(x) = \frac{d}{dx}[-150x^2] + \frac{d}{dx}[6000x] - \frac{d}{dx}[350] = -300x + 6000$
3
Substitute $x = 15$ :
4
$MP(15) = -300(15) + 6000 = 1500$
5
Interpret in context: When 15 bicycles have been sold, profit increases by approximately $1500 for the next additional bicycle sold.

5. Common Pitfalls

Wrong move:

Differentiating $y = 7$ to get $y^{'} = 7$

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 $y = - 4 x^{2}$ to get $y^{'} = - 8 x^{2}$

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 $f (x) = π x$ to get $f^{'} (x) = x + π$

Why:

Students confuse the constant $π$ 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 $\frac{d}{d x} [3 x + 4] = 3 + 4 = 7$

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 $f (x) = x^{3} - 2 x^{2} + x - 4$ to get $f^{'} (x) = 3 x^{2} - 4 x + 1 - 4 = 3 x^{2} - 4 x - 3$

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. Quick Reference Cheatsheet

Rule Name	Formula	Key Notes
Constant Rule	$\frac{d}{d x} [c] = 0$	$c$ is any real constant ( $π, e, k$ included) with no $x$ dependence
Constant Multiple Rule	$\frac{d}{d x} [c \cdot f (x)] = c \cdot f^{'} (x)$	Applies to any constant $c$ (positive/negative) and any differentiable $f (x)$
Sum Rule	$\frac{d}{d x} [f (x) + g (x)] = f^{'} (x) + g^{'} (x)$	Extends to any finite number of terms: derivative of sum = sum of derivatives
Difference Rule	$\frac{d}{d x} [f (x) - g (x)] = f^{'} (x) - g^{'} (x)$	Equivalent to sum rule with $c = - 1$ applied to $g (x)$
Combined Linear Differentiation	$\frac{d}{d x} [c_{1} f (x) + c_{2} g (x)] = c_{1} f^{'} (x) + c_{2} g^{'} (x)$	General form for any linear combination of two differentiable functions
Negative Constant Multiple	$\frac{d}{d x} [- k f (x)] = - k f^{'} (x)$	Always keep the negative sign with the constant when applying the rule

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
Direct differentiation of polynomial
2022 · FRQ
Tangent line slope calculation

Going deeper

unit overviewAP Calculus AB Unit 2 OverviewParent unit for this sub-topic

What's Next

These four rules establish the linearity of differentiation, the foundational property that underpins all differentiation you will do for the rest of AP Calculus AB. Every advanced rule from the product rule to the chain rule relies on this linearity, and common errors here will lead to mistakes on nearly every subsequent differentiation problem. Mastery of this topic is also essential for understanding linearity of integration later in the course, which inherits this property directly from differentiation.