Derivative Rules: Constant, Sum, Difference, and Constant Multiple

AP Calculus · Differentiation: Definition and Fundamental Properties

Constant, Sum, Difference and Constant Multiple Rules

Differentiation is linear — which is exactly why it does not distribute over a product.

key idea$\text{Differentiation is LINEAR: }(af\pm bg)'=af'\pm bg'$

The rule

These four rules are what let you attack a long expression one term at a time. They are easy to state and easy to over-extend — linearity covers sums, differences and constant multiples, and nothing else.

Rule	Statement	In words
Constant	$\dfrac{d}{dx}[c]=0$	a constant never changes, so its rate is zero
Constant multiple	$\dfrac{d}{dx}\left[c\,f(x)\right]=c\,f'(x)$	a constant factor rides along untouched
Sum	$\dfrac{d}{dx}\left[f+g\right]=f'+g'$	differentiate each term separately
Difference	$\dfrac{d}{dx}\left[f-g\right]=f'-g'$	the same, keeping the minus
Product	$\neq f'g'$	NOT linear — needs its own rule
Quotient	$\neq\dfrac{f'}{g'}$	NOT linear — needs its own rule

The last two rows are the point of this table. Linearity is a property of addition, and it does not extend to multiplication or division.

Linear operatorSomething that distributes over addition and lets constants pass through. Differentiation is one; squaring is not, which is why $(a+b)^{2}\neq a^{2}+b^{2}$.

TermA piece separated by $+$ or $-$ at the top level. Terms may be differentiated independently.

FactorA piece joined by multiplication. Factors may not be differentiated independently.

ConstantAny quantity with no $x$ in it — including $\pi$, $e$, $\ln 5$ and an unknown letter such as $k$ when the question says it is fixed.

Pattern recognition

before you split

Split at $+$ and $-$ signs at the top level only.
A constant factor comes out front; a constant term differentiates to $0$.
$\pi$, $e$ and $\ln 2$ are numbers — they behave like $7$.
If pieces are multiplied, stop and use the product rule instead.

Worked examples

EX 1

Term by term

$5x^{4}-3x+8$

Differentiate $f(x)=5x^{4}-3x+8$.

Split into terms

Three terms joined by $-$ and $+$, so each can be handled alone.

First term

Constant multiple plus power rule: $5\cdot4x^{3}=20x^{3}$.

Second term

$-3\cdot1=-3$.

Third term

$8$ is a constant, so it contributes $0$.

Assemble

$f'(x)=20x^{3}-3$.

EX 2

Constants that look like variables

$\pi x^{2}+e^{3}-\ln 5$

Differentiate $g(x)=\pi x^{2}+e^{3}-\ln 5$.

Classify each piece

$\pi$ is a constant factor. $e^{3}$ and $\ln 5$ are constant terms — neither contains $x$.

First term

$\pi\cdot 2x=2\pi x$.

Second and third terms

Both are numbers, so both differentiate to $0$.

Assemble

$g'(x)=2\pi x$.

The trap

$e^{3}$ is a number (about $20.09$), not the function $e^{x}$. Only a variable exponent makes it exponential.

EX 3

Split, then rewrite

$\dfrac{3x^{2}+4\sqrt{x}}{x}$

Differentiate $h(x)=\dfrac{3x^{2}+4\sqrt{x}}{x}$.

Resist the quotient rule

The denominator is a single power, so division distributes over the sum.

Split the fraction

$\dfrac{3x^{2}}{x}+\dfrac{4x^{1/2}}{x}=3x+4x^{-1/2}$.

Differentiate term by term

$3\cdot1=3$ and $4\cdot\left(-\tfrac12\right)x^{-3/2}=-2x^{-3/2}$.

Assemble

$h'(x)=3-2x^{-3/2}=3-\dfrac{2}{x^{3/2}}$.

Why this is legal

Splitting a sum over a common denominator is algebra, not a derivative rule. It converts a quotient into two terms linearity can handle.

EX 4

Where linearity stops

$x^{2}\sin x$

A student writes $\dfrac{d}{dx}\left[x^{2}\sin x\right]=2x\cos x$. Diagnose.

Identify the structure

$x^{2}$ and $\sin x$ are joined by multiplication — they are factors, not terms.

Name the false step

The student differentiated each factor separately, which is linearity applied where it does not hold.

Test the claim quickly

On $f=g=x$: the product is $x^{2}$ with derivative $2x$, but $f'g'=1$. The method fails on the simplest case.

Give the correct answer

By the product rule: $2x\sin x+x^{2}\cos x$.

The discriminator

Ask whether the pieces are added or multiplied. Added ⇒ split freely; multiplied ⇒ use the product rule.

Watch out

Linearity covers $+$ and $-$ only

Products and quotients have their own rules. Applying term-splitting to factors is the single biggest error this topic guards against.

A constant term goes to zero, a constant factor stays

In $5x^{4}$ the $5$ survives as a factor. In $+8$ the $8$ vanishes. Same digit, different role.

$\pi$, $e^{3}$ and $\ln 5$ are numbers

They look like functions but contain no $x$. Only a variable in the expression makes something differentiable-nontrivial.

Splitting a fraction is algebra, not calculus

$\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$ is always true. $\dfrac{a}{b+c}$ does not split — and that is where students over-apply the move.

Course map. Topic 2.6 · LO FUN-3.A — calculate derivatives using the constant, sum, difference and constant multiple rules (EK FUN-3.A.2).