Unit 2.6 – Derivative Rules: Constant, Sum, Difference, and Constant Multiple

AP® Calculus AB & BC | Linearity of Differentiation

Core Concept: Topic 2.6 introduces the four fundamental rules that make differentiation linear! These rules explain how derivatives interact with constants, addition, subtraction, and scalar multiplication. Together, they form the foundation for differentiating ANY polynomial and many complex functions. The beauty: derivatives "play nicely" with basic arithmetic operations—you can break complex expressions into simpler pieces, differentiate each piece, then combine the results! This linearity property is what makes calculus practical and powerful!

🎯 What is Linearity of Differentiation?

LINEARITY OF DIFFERENTIATION

The derivative is a linear operator, which means it satisfies two key properties:

Additivity: The derivative of a sum equals the sum of derivatives
\[ \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)] \]
Homogeneity: Constants can be pulled out of derivatives
\[ \frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)] \]

In Plain English: You can differentiate term-by-term in a sum, and you can "factor out" constant multipliers. This is why polynomials are so easy to differentiate!

📝 Big Picture: The four rules in this topic (Constant, Sum, Difference, Constant Multiple) all stem from the linearity property. They work together to make differentiation systematic and predictable!

📐 The Four Fundamental Rules

RULE 1: THE CONSTANT RULE

The derivative of any constant is zero.

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

Where: \(c\) is any constant (a number without a variable)

Why This Makes Sense:

A constant function \(f(x) = c\) has a horizontal line graph. Horizontal lines have zero slope, so the derivative (rate of change) is zero everywhere!

Examples:

\(\frac{d}{dx}[5] = 0\)
\(\frac{d}{dx}[-12] = 0\)
\(\frac{d}{dx}[\pi] = 0\)
\(\frac{d}{dx}[0] = 0\)
\(\frac{d}{dx}[1000000] = 0\)

Key Takeaway: Constants vanish when you differentiate!

RULE 2: THE CONSTANT MULTIPLE RULE

The derivative of a constant times a function is the constant times the derivative of the function.

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

Where: \(c\) is a constant and \(f(x)\) is a differentiable function

In Words: "Pull the constant out front, then differentiate."

Examples:

\(\frac{d}{dx}[7x^3] = 7 \cdot \frac{d}{dx}[x^3] = 7 \cdot 3x^2 = 21x^2\)
\(\frac{d}{dx}[-4x^5] = -4 \cdot 5x^4 = -20x^4\)
\(\frac{d}{dx}[\frac{1}{2}x^2] = \frac{1}{2} \cdot 2x = x\)
\(\frac{d}{dx}[\pi x] = \pi \cdot 1 = \pi\)

Common Application: When differentiating \(ax^n\), use constant multiple rule THEN power rule:

\[ \frac{d}{dx}[ax^n] = a \cdot nx^{n-1} = anx^{n-1} \]

RULE 3: THE SUM RULE

The derivative of a sum is the sum of the derivatives.

\[ \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)] = f'(x) + g'(x) \]

In Words: "Differentiate each term separately, then add the results."

Extends to Multiple Terms:

\[ \frac{d}{dx}[f(x) + g(x) + h(x)] = f'(x) + g'(x) + h'(x) \]

Examples:

\(\frac{d}{dx}[x^3 + x^2] = 3x^2 + 2x\)
\(\frac{d}{dx}[x^5 + 2x] = 5x^4 + 2\)
\(\frac{d}{dx}[x^4 + x^3 + x^2 + x] = 4x^3 + 3x^2 + 2x + 1\)
\(\frac{d}{dx}[\sqrt{x} + \frac{1}{x}] = \frac{1}{2\sqrt{x}} - \frac{1}{x^2}\)

Why It Works: The limit of a sum equals the sum of the limits!

RULE 4: THE DIFFERENCE RULE

The derivative of a difference is the difference of the derivatives.

\[ \frac{d}{dx}[f(x) - g(x)] = \frac{d}{dx}[f(x)] - \frac{d}{dx}[g(x)] = f'(x) - g'(x) \]

In Words: "Differentiate each term separately, then subtract."

Note: This is really just the Sum Rule with a negative constant multiple:

\[ f(x) - g(x) = f(x) + (-1) \cdot g(x) \]

Examples:

\(\frac{d}{dx}[x^3 - x^2] = 3x^2 - 2x\)
\(\frac{d}{dx}[5x^4 - 2x] = 20x^3 - 2\)
\(\frac{d}{dx}[x^2 - 7] = 2x - 0 = 2x\) (constant disappears!)
\(\frac{d}{dx}[\frac{1}{x} - \sqrt{x}] = -\frac{1}{x^2} - \frac{1}{2\sqrt{x}}\)

⚠️ Watch Signs: Keep track of negative signs carefully!

🔗 Combining All Four Rules

The Power of Linearity: Using All Rules Together

For ANY polynomial (or linear combination of differentiable functions), you can:

Break it into terms using Sum/Difference Rules
Pull out constants using Constant Multiple Rule
Apply Power Rule (or other derivative formulas) to each term
Constants disappear using Constant Rule
Combine results using Sum/Difference Rules again

General Formula for Polynomials:

\[ \frac{d}{dx}[a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0] = na_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 \]

📖 Comprehensive Worked Examples

Example 1: Basic Polynomial

Problem: Find \(\frac{d}{dx}[3x^4 - 5x^2 + 7x - 2]\)

Solution (Using All Four Rules):

Sum/Difference Rule: Break into terms
\[ \frac{d}{dx}[3x^4] - \frac{d}{dx}[5x^2] + \frac{d}{dx}[7x] - \frac{d}{dx}[2] \]
Constant Multiple Rule: Pull out constants
\[ 3\frac{d}{dx}[x^4] - 5\frac{d}{dx}[x^2] + 7\frac{d}{dx}[x] - \frac{d}{dx}[2] \]
Power Rule & Constant Rule: Differentiate each
- \(3 \cdot 4x^3 = 12x^3\)
- \(5 \cdot 2x = 10x\)
- \(7 \cdot 1 = 7\)
- \(0\) (constant rule)
Combine: \(12x^3 - 10x + 7\)

Answer: \(12x^3 - 10x + 7\)

Example 2: With Fractions

Problem: Find \(f'(x)\) if \(f(x) = \frac{2}{3}x^6 - \frac{1}{4}x^3 + \frac{5}{2}\)

Solution:

Separate terms:
\[ f'(x) = \frac{d}{dx}\left[\frac{2}{3}x^6\right] - \frac{d}{dx}\left[\frac{1}{4}x^3\right] + \frac{d}{dx}\left[\frac{5}{2}\right] \]
Apply rules:
- \(\frac{2}{3} \cdot 6x^5 = 4x^5\)
- \(\frac{1}{4} \cdot 3x^2 = \frac{3}{4}x^2\)
- \(0\) (constant)
Result: \(f'(x) = 4x^5 - \frac{3}{4}x^2\)

Answer: \(f'(x) = 4x^5 - \frac{3}{4}x^2\)

Example 3: Long Polynomial

Problem: Differentiate \(g(x) = x^5 + 3x^4 - 2x^3 + 7x^2 - 5x + 10\)

Solution (Term-by-Term):

\[ g'(x) = 5x^4 + 12x^3 - 6x^2 + 14x - 5 \]

Breakdown:

\(\frac{d}{dx}[x^5] = 5x^4\)
\(\frac{d}{dx}[3x^4] = 12x^3\)
\(\frac{d}{dx}[-2x^3] = -6x^2\)
\(\frac{d}{dx}[7x^2] = 14x\)
\(\frac{d}{dx}[-5x] = -5\)
\(\frac{d}{dx}[10] = 0\)

Answer: \(g'(x) = 5x^4 + 12x^3 - 6x^2 + 14x - 5\)

Example 4: Mixed Exponents

Problem: Find \(\frac{dy}{dx}\) for \(y = 4x^3 - \frac{2}{x^2} + 3\sqrt{x} - 7\)

Solution:

Rewrite with exponents:
\[ y = 4x^3 - 2x^{-2} + 3x^{1/2} - 7 \]
Differentiate term-by-term:
- \(\frac{d}{dx}[4x^3] = 12x^2\)
- \(\frac{d}{dx}[-2x^{-2}] = -2(-2)x^{-3} = 4x^{-3}\)
- \(\frac{d}{dx}[3x^{1/2}] = 3 \cdot \frac{1}{2}x^{-1/2} = \frac{3}{2}x^{-1/2}\)
- \(\frac{d}{dx}[-7] = 0\)
Combine:
\[ \frac{dy}{dx} = 12x^2 + 4x^{-3} + \frac{3}{2}x^{-1/2} \]
Rewrite (optional):
\[ \frac{dy}{dx} = 12x^2 + \frac{4}{x^3} + \frac{3}{2\sqrt{x}} \]

Answer: \(\frac{dy}{dx} = 12x^2 + \frac{4}{x^3} + \frac{3}{2\sqrt{x}}\)

Example 5: Factored Form (Must Expand First)

Problem: Find \(h'(x)\) for \(h(x) = (x + 2)(x - 3)\)

Solution:

Expand first:
\[ h(x) = x^2 - 3x + 2x - 6 = x^2 - x - 6 \]
Now differentiate:
\[ h'(x) = 2x - 1 \]

Answer: \(h'(x) = 2x - 1\)

Note: You could use the Product Rule (Topic 2.8), but expanding first is usually simpler for polynomials!

Example 6: Combining Like Terms

Problem: Differentiate \(f(x) = 2x^3 + 5x^3 - x^2 + 4x^2\)

Solution:

Option 1: Simplify first (recommended)
- Combine: \(f(x) = 7x^3 + 3x^2\)
- Differentiate: \(f'(x) = 21x^2 + 6x\)
Option 2: Differentiate then simplify
- \(f'(x) = 6x^2 + 15x^2 - 2x + 8x\)
- Combine: \(f'(x) = 21x^2 + 6x\)

Answer: \(f'(x) = 21x^2 + 6x\)

Tip: Simplifying BEFORE differentiating saves work!

📐 Why These Rules Work: Proofs

Proof of the Sum Rule

Theorem: If \(f\) and \(g\) are differentiable, then \(\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\)

Proof:

Start with limit definition:
\[ \frac{d}{dx}[f(x) + g(x)] = \lim_{h \to 0} \frac{[f(x+h) + g(x+h)] - [f(x) + g(x)]}{h} \]
Rearrange numerator:
\[ = \lim_{h \to 0} \frac{[f(x+h) - f(x)] + [g(x+h) - g(x)]}{h} \]
Separate into two fractions:
\[ = \lim_{h \to 0} \left[\frac{f(x+h) - f(x)}{h} + \frac{g(x+h) - g(x)}{h}\right] \]
Use limit law (limit of sum = sum of limits):
\[ = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} + \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \]
Recognize derivatives:
\[ = f'(x) + g'(x) \quad \checkmark \]

Proof of the Constant Multiple Rule

Theorem: If \(f\) is differentiable and \(c\) is a constant, then \(\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\)

Proof:

Start with limit definition:
\[ \frac{d}{dx}[c \cdot f(x)] = \lim_{h \to 0} \frac{c \cdot f(x+h) - c \cdot f(x)}{h} \]
Factor out constant \(c\) from numerator:
\[ = \lim_{h \to 0} \frac{c[f(x+h) - f(x)]}{h} \]
Use limit law (constant can be pulled out):
\[ = c \cdot \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Recognize derivative:
\[ = c \cdot f'(x) \quad \checkmark \]

📝 Note: The Difference Rule and Constant Rule follow easily from these two proofs. The Difference Rule is just Sum Rule with \(g(x)\) replaced by \(-g(x)\). The Constant Rule comes from recognizing that a constant \(c\) can be written as \(c \cdot x^0\), and its derivative is \(c \cdot 0 \cdot x^{-1} = 0\).

📋 Complete Summary of All Four Rules

Rule Name	Formula	In Words	Example
Constant Rule	\(\frac{d}{dx}[c] = 0\)	Derivative of constant is zero	\(\frac{d}{dx}[5] = 0\)
Constant Multiple	\(\frac{d}{dx}[cf] = c \cdot f'\)	Pull constant out, then differentiate	\(\frac{d}{dx}[7x^2] = 14x\)
Sum Rule	\(\frac{d}{dx}[f + g] = f' + g'\)	Derivative of sum = sum of derivatives	\(\frac{d}{dx}[x^2 + x] = 2x + 1\)
Difference Rule	\(\frac{d}{dx}[f - g] = f' - g'\)	Derivative of difference = difference of derivatives	\(\frac{d}{dx}[x^3 - x] = 3x^2 - 1\)

💡 Tips, Tricks & Strategies

✅ Essential Tips

Always simplify first: Combine like terms before differentiating to save work
Expand products: For simple polynomials, expand before differentiating (Product Rule not needed yet)
Constants vanish: Any standalone number becomes zero when differentiated
Work term-by-term: Break complex expressions into pieces, differentiate each, then recombine
Keep track of signs: Negative signs stay with their terms through differentiation
Fractional coefficients: Just multiply them by the power rule exponent
Check your work: Count terms—if original had 5 terms and one was constant, derivative should have 4

🎯 Step-by-Step Workflow

Universal Process for Differentiating Any Expression:

REWRITE: Convert all roots and fractions to exponent form
EXPAND: Multiply out any products (if simple enough)
COMBINE: Add/subtract like terms to simplify
SEPARATE: Identify each term (use Sum/Difference Rules)
DIFFERENTIATE: Apply Constant Multiple Rule + Power Rule to each term
SIMPLIFY: Combine results and clean up notation
VERIFY: Quick sanity check (does degree drop by 1?)

🔥 Memory Aids

Acronym for the Four Rules:

"CSCD" - Constant, Sum, Constant Multiple, Difference

Rhyme to Remember:

"Constants are zero, pull numbers out,
Add derivatives when adding throughout.
Subtract derivatives when you subtract—
Linearity makes differentiation exact!"

❌ Common Mistakes to Avoid

Mistake 1: Thinking \(\frac{d}{dx}[5x] = 0\) because "constants are zero" → NO! It's \(5\) (constant times x)
Mistake 2: Forgetting to distribute negative signs: \(\frac{d}{dx}[x^2 - 3x]\) needs BOTH derivatives subtracted
Mistake 3: Trying to use these rules on products like \((x^2)(x^3)\) → Must expand to \(x^5\) first OR use Product Rule
Mistake 4: Losing constant multiples: \(\frac{d}{dx}[6x^3] = 18x^2\), not \(3x^2\)
Mistake 5: Not simplifying final answer (leave \(2x + 3x\) as \(5x\))
Mistake 6: Thinking Sum Rule applies to products → \(\frac{d}{dx}[f \cdot g] \neq f' \cdot g'\) (wrong!)
Mistake 7: Forgetting that \(x\) has an invisible coefficient of 1
Mistake 8: Applying rules to composite functions (need Chain Rule from 2.9)

📝 Practice Problems

Find the derivative of each function:

\(f(x) = 8x^3 + 5x^2 - 3x + 7\)
\(g(x) = -4x^5 + 2x^3 - x + 12\)
\(h(x) = \frac{3}{4}x^4 - \frac{2}{3}x^2 + \frac{1}{2}\)
\(k(x) = 100x^{10} - 50x^5 + 25x^2 - 10\)
\(f(x) = (x - 1)(x + 3)\) (expand first)
\(g(x) = x^2 + 2x^2 - 5 + 3\) (simplify first)

Answers:

\(f'(x) = 24x^2 + 10x - 3\)
\(g'(x) = -20x^4 + 6x^2 - 1\)
\(h'(x) = 3x^3 - \frac{4}{3}x\)
\(k'(x) = 1000x^9 - 250x^4 + 50x\)
Expand: \(f(x) = x^2 + 2x - 3\), so \(f'(x) = 2x + 2\)
Simplify: \(g(x) = 3x^2 - 2\), so \(g'(x) = 6x\)

✏️ AP® Exam Success Tips

What the AP® Exam Expects:

Show rule application: On FRQ, briefly indicate which rules you're using
Simplify completely: Combine like terms in final answer
Work systematically: Go term-by-term to avoid missing anything
Check signs carefully: Negative terms stay negative
Don't skip steps: Show differentiation of each term on FRQ
Know when rules DON'T apply: These are for sums/differences only, not products or quotients
Multiple representations OK: \(\frac{3}{2}x\) and \(1.5x\) are both acceptable

Common FRQ Formats:

"Find f'(x) for the function..." (straightforward differentiation)
"Show that the derivative of f(x) = ... is ..." (must show work)
"Find the slope of the tangent line..." (evaluate f' at specific point)
"Determine the rate of change..." (find and evaluate derivative)
"For what values of x is f'(x) = ..." (solve equation after finding derivative)
"Find an equation of the tangent line..." (need f'(a) as slope)

⚡ Quick Reference Card

THE FOUR LINEARITY RULES

Rule	Formula	Quick Guide
Constant	\(\frac{d}{dx}[c] = 0\)	Numbers → 0
Constant Multiple	\(\frac{d}{dx}[cf] = c \cdot f'\)	Pull out, differentiate
Sum	\(\frac{d}{dx}[f + g] = f' + g'\)	Add derivatives
Difference	\(\frac{d}{dx}[f - g] = f' - g'\)	Subtract derivatives

Together these establish: DIFFERENTIATION IS LINEAR!

🔗 Why This Topic Matters

Topic 2.6 connects to:

Topic 2.5: These rules are used WITH the Power Rule constantly
Topics 2.7-2.9: All derivative rules (trig, exponential, product, quotient, chain) build on linearity
Unit 3: Finding critical points requires polynomial derivatives
Unit 4: Motion problems use derivatives of position functions
Unit 5: Integration is "reverse" of differentiation
All calculus: Linearity is WHY calculus is computationally feasible!

Remember: The four derivative rules in Topic 2.6 establish the linearity of differentiation: (1) Constant Rule: \(\frac{d}{dx}[c] = 0\)—constants vanish; (2) Constant Multiple Rule: \(\frac{d}{dx}[cf] = c \cdot f'\)—pull constants out; (3) Sum Rule: \(\frac{d}{dx}[f + g] = f' + g'\)—add derivatives; (4) Difference Rule: \(\frac{d}{dx}[f - g] = f' - g'\)—subtract derivatives. Together, these rules mean you can differentiate term-by-term in any sum or difference, making polynomials and linear combinations trivial to differentiate! Combined with the Power Rule from 2.5, you now have complete mastery of polynomial differentiation. Master these four rules—they're the foundation of ALL derivative calculations! 🎯✨