Sum, Difference, and Constant Multiple Rules

MATH161

Reference: Stewart §2.3 • Chapter: 2 • Section: 3

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Sum, Difference, and Constant Multiple Rules

Differentiation is a Linear Operation

Here's a powerful idea: differentiation "plays nicely" with addition and scalar multiplication. If you can differentiate individual pieces, you can differentiate their sum just by adding up the derivatives. This means that once you know the Power Rule, you can differentiate any polynomial—no matter how complicated.

These rules capture a fundamental property: the derivative operator is linear. In mathematical notation, $\frac{d}{dx}$ distributes over sums just like multiplication does.

Prerequisite Map

Quick Reference

Property	Value
Concept	Differentiation Formulas
Chapter	2.3
Difficulty	Beginner
Time	~15 minutes

Key Concepts

The Three Rules

Constant Multiple Rule: If $c$ is a constant and $f$ is differentiable, then: $$\boxed{\frac{d}{dx}[cf(x)] = c\frac{d}{dx}[f(x)] = cf'(x)}$$

"Constants factor out of derivatives."

Sum Rule: If $f$ and $g$ are both differentiable, then: $$\boxed{\frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)] = f'(x) + g'(x)}$$

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

Difference Rule: If $f$ and $g$ are both differentiable, then: $$\boxed{\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)}$$

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

Using Prime Notation

These rules are compact in prime notation:

Rule	Leibniz Notation	Prime Notation
Constant Multiple	$\frac{d}{dx}[cf] = cf'$	$(cf)' = cf'$
Sum	$\frac{d}{dx}[f+g] = f' + g'$	$(f+g)' = f' + g'$
Difference	$\frac{d}{dx}[f-g] = f' - g'$	$(f-g)' = f' - g'$

Why These Rules Work

Constant Multiple Rule Proof:

Let $g(x) = cf(x)$. Then: $$g'(x) = \lim_{h \to 0}\frac{g(x+h) - g(x)}{h} = \lim_{h \to 0}\frac{cf(x+h) - cf(x)}{h}$$ $$= \lim_{h \to 0}c \cdot \frac{f(x+h) - f(x)}{h} = c \cdot \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = cf'(x)$$

The constant $c$ pulls out of the limit because it doesn't depend on $h$.

Sum Rule Proof:

Let $F(x) = f(x) + g(x)$. Then: $$F'(x) = \lim_{h \to 0}\frac{[f(x+h) + g(x+h)] - [f(x) + g(x)]}{h}$$ $$= \lim_{h \to 0}\left[\frac{f(x+h) - f(x)}{h} + \frac{g(x+h) - g(x)}{h}\right] = f'(x) + g'(x)$$

Combining with the Power Rule

These rules combine with the Power Rule to differentiate any polynomial:

Example: Differentiate $p(x) = 3x^4 - 5x^2 + 7x - 2$

Step by step: $$p'(x) = \frac{d}{dx}(3x^4) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(7x) - \frac{d}{dx}(2)$$ $$= 3 \cdot 4x^3 - 5 \cdot 2x + 7 \cdot 1 - 0$$ $$= 12x^3 - 10x + 7$$

Extension to Multiple Terms

The Sum Rule extends to any number of functions: $$(f + g + h)' = f' + g' + h'$$

This follows by applying the Sum Rule twice: $$(f + g + h)' = ((f + g) + h)' = (f + g)' + h' = f' + g' + h'$$

Practice Problems

Level 1 Constant Multiple

Differentiate $f(x) = 7x^5$.

Thought Process

The 7 is a constant, so it stays in front. Apply the Power Rule to $x^5$, then multiply by 7.

Show Answer

$$f'(x) = 7 \cdot \frac{d}{dx}(x^5) = 7 \cdot 5x^4 = 35x^4$$

Level 2 Simple Sum

Find $\frac{dy}{dx}$ if $y = x^3 + x^2 - 4x$.

Thought Process

Use the Sum/Difference Rule to differentiate term by term. Apply the Power Rule to each term: $x^3$, $x^2$, and $x$.

Show Answer

$$\frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2) - \frac{d}{dx}(4x)$$ $$= 3x^2 + 2x - 4$$

Level 3 Full Polynomial

Differentiate $g(t) = 2t^6 - 3t^4 + 5t^2 - 8t + 11$.

Thought Process

Apply the rules term by term. For each term like $2t^6$, first factor out the constant (2), then apply the Power Rule to $t^6$, then multiply.

Show Answer

Term by term:

$\frac{d}{dt}(2t^6) = 2 \cdot 6t^5 = 12t^5$
$\frac{d}{dt}(-3t^4) = -3 \cdot 4t^3 = -12t^3$
$\frac{d}{dt}(5t^2) = 5 \cdot 2t = 10t$
$\frac{d}{dt}(-8t) = -8$
$\frac{d}{dt}(11) = 0$

Combining: $$g'(t) = 12t^5 - 12t^3 + 10t - 8$$

Level 4 Mixed Exponents

Find $f'(x)$ if $f(x) = 4\sqrt{x} - \frac{3}{x^2} + 2x$.

Thought Process

First convert to exponential form: $\sqrt{x} = x^{1/2}$ and $\frac{1}{x^2} = x^{-2}$. Then apply the sum rule and power rule to each term.

Show Answer

Rewrite: $f(x) = 4x^{1/2} - 3x^{-2} + 2x$

Differentiate term by term: $$f'(x) = 4 \cdot \frac{1}{2}x^{-1/2} - 3 \cdot (-2)x^{-3} + 2$$ $$= 2x^{-1/2} + 6x^{-3} + 2$$ $$= \frac{2}{\sqrt{x}} + \frac{6}{x^3} + 2$$

Level 5 Finding Horizontal Tangents

Find all points on the curve $y = x^3 - 6x^2 + 9x + 2$ where the tangent line is horizontal.

Thought Process

A horizontal tangent has slope 0. Find $y'$, set it equal to 0, solve for $x$, then find the corresponding $y$-values.

Show Answer

Step 1: Find the derivative. $$y' = 3x^2 - 12x + 9$$

Step 2: Set $y' = 0$ (horizontal tangent means slope = 0). $$3x^2 - 12x + 9 = 0$$ $$3(x^2 - 4x + 3) = 0$$ $$3(x - 1)(x - 3) = 0$$

So $x = 1$ or $x = 3$.

Step 3: Find the $y$-coordinates.

When $x = 1$: $y = 1 - 6 + 9 + 2 = 6$
When $x = 3$: $y = 27 - 54 + 27 + 2 = 2$

Answer: The curve has horizontal tangents at $(1, 6)$ and $(3, 2)$.

CCI-Style Conceptual Questions

Conceptual Interpreting the Rules

If $f'(2) = 5$ and $g'(2) = -3$, find $(3f - 2g)'(2)$.

Thought Process

Apply the Constant Multiple and Difference Rules: $(3f - 2g)' = 3f' - 2g'$. Then substitute the known values.

Show Answer

By the Constant Multiple and Difference Rules: $$(3f - 2g)'(2) = 3f'(2) - 2g'(2) = 3(5) - 2(-3) = 15 + 6 = 21$$

Conceptual Parallel Tangent Lines

Curves $y = 2x^2$ and $y = x^2 + k$ have parallel tangent lines at every value of $x$. What does this tell you about $k$?

Thought Process

For tangent lines to be parallel at every $x$, the derivatives must be equal everywhere. Compute both derivatives and compare.

Show Answer

Compute derivatives:

$y = 2x^2 \Rightarrow y' = 4x$
$y = x^2 + k \Rightarrow y' = 2x$

These are NOT equal for all $x$ (unless $x = 0$). So the tangent lines are NOT parallel everywhere.

Wait—the premise is false! The curves $y = 2x^2$ and $y = x^2 + k$ do NOT have parallel tangent lines at every $x$. The constant $k$ only shifts the curve vertically; it doesn't affect the slopes.

For parallel tangents everywhere, we'd need curves like $y = x^2$ and $y = x^2 + k$, which have the same derivative $y' = 2x$.

Mastery Checklist

[ ] I can differentiate a constant times a function
[ ] I can differentiate sums and differences term by term
[ ] I can combine these rules to differentiate any polynomial
[ ] I can handle negative and fractional exponents in combinations
[ ] I can find where a polynomial has horizontal tangents

Mental Model

Think of derivatives as "linear operators":

Just as distributing multiplication over addition works: $$3(a + b) = 3a + 3b$$

Differentiation distributes over addition too: $$\frac{d}{dx}(f + g) = \frac{d}{dx}f + \frac{d}{dx}g$$

The derivative "respects" the structure of linear combinations. This is why polynomials are so easy to differentiate—you just handle each term separately.

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Last updated: 2026-01-22