Sum, difference, constant multiple rules: Question bank

Definition The vocabulary this page runs on

A few words first, so nothing is used before it is built.

Derivative $f^{'} (x)$ — the slope of the curve $y = f (x)$ at the point $x$ : how steep the graph is right there.
Difference quotient — the slope of the little line joining two nearby points on the curve, one at $x$ and one a tiny step further along. That tiny horizontal step we call == $Δ x$ == (read "delta-x"), just a small change in $x$ . The difference quotient is $\frac{f ( x + Δ x ) - f ( x )}{Δ x}$ (rise over run). Shrinking $Δ x$ toward $0$ turns this line into the true slope — that is the derivative.
Linear — a rule is linear if it does all three of these:
1. splits across a sum: $[f + g]^{'} = f^{'} + g^{'}$
2. splits across a difference: $[f - g]^{'} = f^{'} - g^{'}$
3. lets a constant factor slide out: $[c f]^{'} = c f^{'}$

(The ==...== marks are just this vault's way of flagging a key term for flashcards — treat them as a highlighter, not maths.)

The two pictures below are the entire engine behind these rules. Every trap on the page is really just one of these pictures misread — glance back at them whenever an answer says "the slopes add" or "the constant rides along".

Figure — Sum, difference, constant multiple rules

True or false — justify

The derivative of a sum is the sum of the derivatives, for any two differentiable functions.

True — this is exactly the sum rule

\frac{d}{d x} [f + g] = f^{'} + g^{'}

. Picture s01: the combined curve's rise over any step is just

f

's rise plus

g

's rise, so the slopes stack; formally the sum law for limits lets the difference quotient split into two.

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

False — this is the classic over-generalisation. Linearity covers

+, -, \times constant

only; products need the Product rule

(f g)^{'} = f^{'} g + f g^{'}

. Test:

\frac{d}{d x} (x \cdot x) = 2 x

but

x^{'} \cdot x^{'} = 1

\frac{d}{d x} [5 f (x)] = 5 f^{'} (x)

for every constant and every differentiable

f

True — the constant multiple rule; the

5

never depends on

Δ x

(the tiny step), so in the difference quotient it factors out and rides straight through the limit unchanged.

\frac{d}{d x} [f (x) \cdot 5] = f^{'} (x) \cdot 5

is a different rule from the constant multiple rule.

False — it is the same rule. Multiplying by a constant is commutative, so

5 f

and

f \cdot 5

are identical, and both give

5 f^{'}

c

depends on

x

— say

c = x

— then

\frac{d}{d x} [c f] = c f^{'}

still holds.

False — "constant" must literally mean constant. If

c = x

then

\frac{d}{d x} [x f]

needs the product rule, giving

f + x f^{'}

, not

x f^{'}

The derivative of

f (x) - g (x)

requires its own separate proof from the limit definition.

False — since

f - g = f + (- 1) g

, the sum rule plus the constant multiple rule (with

c = - 1

) hand it to you for free.

Linearity works for three or more terms at once, e.g.

\frac{d}{d x} [f + g + h]

True — apply the sum rule repeatedly (group as

(f + g) + h

). Linearity extends to any finite number of terms with any scalar coefficients.

\frac{d}{d x} [2 sin x + 3 e^{x}]

can be computed without ever going back to the difference quotient.

True — that is the whole payoff of linearity: differentiate each known piece and recombine, no limits needed at the point of use.

Spot the error

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

." — find the mistake.

The derivative of the constant

4

0

, not

4

. Correct:

2 x - 3

. A constant is a flat line, so its slope is zero.

\frac{d}{d x} [f - (x^{2} - 3 x)] = f^{'} - 2 x - 3

." — find the mistake.

The minus sign must distribute over the whole bracket. Treat it as

(- 1) (x^{2} - 3 x)

: derivative is

- (2 x - 3) = - 2 x + 3

, so the answer is

f^{'} - 2 x + 3

\frac{d}{d x} [x^{2} \cdot sin x] = 2 x \cdot cos x

." — find the mistake.

Products don't split. This wrongly applied the sum-style rule to a product. Use the product rule:

2 x sin x + x^{2} cos x

\frac{d}{d x} [\frac{x}{5}] = \frac{1}{5 x}

." — find the mistake.

\frac{x}{5}

\frac{1}{5} x

, a constant

\frac{1}{5}

times

x

, so its derivative is

\frac{1}{5} \cdot 1 = \frac{1}{5}

. The student confused it with

\frac{1}{5 x} = \frac{1}{5} x^{- 1}

\frac{d}{d x} [7 f (x)] = 7 + f^{'} (x)

." — find the mistake.

A constant multiplies

f

, it does not add to it. The constant multiple rule gives

7 f^{'} (x)

, not

7 + f^{'} (x)

"Since

\frac{d}{d x} [c f] = c f^{'}

, then

\frac{d}{d x} [c + f] = c + f^{'}

." — find the mistake.

The added constant

c

has derivative

0

, so

\frac{d}{d x} [c + f] = f^{'}

, not

c + f^{'}

. Adding a constant shifts the graph up but never tilts it, so the slope is unchanged.

Why questions

Why does the sum rule fall out of the limit definition but the product rule does not?

Look at s01: the numerator

[f (x + Δ x) + g (x + Δ x)] - [f (x) + g (x)]

regroups into an

f

-difference plus a

g

-difference — two clean difference quotients. A product

f (x + Δ x) g (x + Δ x)

instead leaves a cross-term that refuses to separate, forcing the extra

f^{'} g + f g^{'}

structure.

Why is it legitimate to pull a constant

c

outside the limit in the constant multiple proof?

Because

c

does not depend on

Δ x

(the shrinking step), so as

Δ x \to 0

it stays fixed; the limit law "constant factors pull out of limits" applies exactly. In s02 the whole curve is stretched vertically by

c

, so every slope is scaled by

c

too.

Why must

f

and

g

each be differentiable for the sum rule to hold?

The sum law for limits needs both limits (both slopes) to exist before you can split them. If either

f^{'}

g^{'}

fails to exist, the split is not justified — even though the sum might still be differentiable in special cases (see the last edge case below).

Why is calling

\frac{d}{d x}

a "linear operator" more than just a slogan?

Because it satisfies the two defining properties of linearity — additivity and scalar homogeneity — captured in

\frac{d}{d x} [a f + b g] = a f^{'} + b g^{'}

. This places it alongside other Linear operators and mirrors Linearity of integration.

Why does the difference rule not need its own limit-law?

Difference is just addition of a negative:

f - g = f + (- 1) g

. So the same two ingredients (sum law + constant pulls out, both pictured in s01 and s02) already cover it — subtraction introduces nothing new.

Why can linearity mix a sine, an exponential, and a logarithm in one derivative without any interaction between them?

Because each term is a scalar times a function that is added to the others; linearity treats each summand independently (each is its own layer in the s01 stack). Interaction only appears when functions are multiplied or composed, which linearity does not govern.

Edge cases

What is

\frac{d}{d x} [0 \cdot f (x)]

It is

0 \cdot f^{'} (x) = 0

— a valid instance of the constant multiple rule with

c = 0

. The function is flattened onto the

x

-axis, so its slope is zero everywhere.

What is

\frac{d}{d x} [f (x) - f (x)]

The difference rule gives

f^{'} (x) - f^{'} (x) = 0

, consistent with the fact that

f - f

is the constant function

0

. Both routes agree — a nice sanity check.

Does linearity say anything about

\frac{d}{d x} [f (x) /2]

versus

\frac{d}{d x} [2/ f (x)]

The first is

\frac{1}{2} f^{'} (x)

(constant multiple, since

/2 = \times \frac{1}{2}

). The second has the variable

f (x)

in the denominator, so it is not a constant multiple and needs the quotient/chain rules instead.

g (x)

is differentiable but

f (x)

is not, can

\frac{d}{d x} [f + g]

still exist?

Yes, in principle the sum could be differentiable even when one piece is not — but the sum rule does not apply, because it requires both slopes to exist to split the limit. You must analyse the sum directly.

Two functions that are each non-differentiable can sum to a smooth function — how? (See figure.)

Take

f (x) = ∣ x ∣

and

g (x) = - ∣ x ∣

: each has a sharp corner at

x = 0

where no single slope exists. But

f (x) + g (x) = 0

is a flat line, perfectly differentiable with derivative

0

. Picture s03: the two corners are mirror images and cancel. The lesson — you may not invoke the sum rule here (neither

f^{'} (0)

nor

g^{'} (0)

exists), even though the sum is differentiable. Always check both pieces first.

What does the constant multiple rule give for a negative constant, e.g.

\frac{d}{d x} [- 4 f (x)]

It gives

- 4 f^{'} (x)

; the sign rides along just like any other constant. This is exactly the mechanism (

c = - 1

) that generates the difference rule.

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

a special case of the constant multiple rule?

Yes — write the constant

c

c \cdot 1

where the function is the constant

1

; then

c \cdot (1)^{'} = c \cdot 0 = 0

. A flat line has zero slope in every case.

Connections

Sum, difference, constant multiple rules — parent note with the full proofs.
Limit laws (sum, scalar, product of limits) — justifies every split above.
Definition of the derivative (limit of difference quotient) — the ground floor.
Power rule — pairs with linearity to differentiate any polynomial.
Product rule — the non-linear sibling these traps keep pointing at.
Linearity of integration · Linear operators — the same structure, elsewhere.