Sum, difference, constant multiple rules: Exercises

Intuition How to use this page

Every problem below tests one idea from Sum, difference, constant multiple rules: differentiation is a linear operation — slopes add, constants ride along, and subtraction is just adding a negative. Work each problem before opening its solution. The difficulty climbs in five levels; a trap-buster [!mistake] closes each level so you learn where the cliff edge is.

The only tools you are allowed here, all built in the parent note:

Definition of the derivative — see Definition of the derivative (limit of difference quotient).
The three linearity rules: $\frac{d}{d x} [f + g] = f^{'} + g^{'}$ , $\frac{d}{d x} [f - g] = f^{'} - g^{'}$ , $\frac{d}{d x} [c f] = c f^{'}$ .
The Power rule: $\frac{d}{d x} x^{n} = n x^{n - 1}$ (for any real $n$ ).
A few known single-function derivatives: $sin^{'} = cos$ , $cos^{'} = - sin$ , $(e^{x})^{'} = e^{x}$ , $ln^{'} = 1/ x$ .

Before we start, one figure to fix the meaning of " $^{'}$ ": it is a slope, the steepness of the curve at a point. Everything on this page is bookkeeping on slopes.

Figure — Sum, difference, constant multiple rules

Look at the red tangent line: its steepness is the number $f^{'} (x)$ . When we "add slopes," we are literally adding the steepness of two such red lines.

Level 1 — Recognition

Here you only need to spot which rule applies and read off the answer. No algebra tricks.

Recall Solution 1.1

WHAT: A sum of two powers. WHY that rule: the sum rule says split the sum, then power rule each piece. $y^{'} = \frac{d}{d x} x^{7} + \frac{d}{d x} x^{3} = 7 x^{6} + 3 x^{2} .$

Recall Solution 1.2

WHAT: A constant $5$ times a power. WHY: constant multiple rule pulls the $5$ out; power rule handles $x^{2}$ . $y^{'} = 5 \cdot \frac{d}{d x} x^{2} = 5 \cdot 2 x = 10 x .$

Recall Solution 1.3

WHAT: A scaled power minus a constant. $y^{'} = 4 \cdot 3 x^{2} - \frac{d}{d x} (9) = 12 x^{2} - 0 = 12 x^{2} .$ The constant $9$ is a flat line — zero slope — so it contributes $0$ .

Level 2 — Application

Now you apply several rules in one problem, including a rewrite before you can see the power.

Recall Solution 2.1

Split every term (sum/difference), pull out constants, power-rule each: $y^{'} = 3 \cdot 4 x^{3} - 5 \cdot 2 x + 6 \cdot 1 - 0 = 12 x^{3} - 10 x + 6.$

Recall Solution 2.2

Step 1 — rewrite as powers. WHY: the power rule needs the form $x^{n}$ ; a fraction and a root are hidden powers. $\frac{2}{x} = 2 x^{- 1}, x = x^{1/2} .$ Step 2 — differentiate. $y^{'} = 2 \cdot (- 1) x^{- 2} + \frac{1}{2} x^{- 1/2} = - \frac{2}{x ^{2}} + \frac{1}{2 x} .$

Recall Solution 2.3

Linearity lets the trig terms coexist without interacting. Recall $sin^{'} = cos$ , $cos^{'} = - sin$ : $f^{'} (x) = 2 cos x - 3 (- sin x) + 0 = 2 cos x + 3 sin x .$ Note the double sign flip on the cosine term: subtracting $3 cos x$ and $cos^{'} = - sin$ turns into $+ 3 sin x$ .

Level 3 — Analysis

Here signs and grouping bite. You must reason about how the minus distributes.

Recall Solution 3.1

WHAT: A subtraction of a whole multi-term function $g (x) = 2 x^{2} - 5 x + 1$ . WHY care: the minus is really $(- 1) \cdot g$ , so it multiplies every term of $g$ . First distribute conceptually: $y = x^{3} - 2 x^{2} + 5 x - 1$ . $y^{'} = 3 x^{2} - 4 x + 5 - 0 = 3 x^{2} - 4 x + 5.$

Recall Solution 3.2

Step 1 — rewrite as scaled powers. Dividing by $2$ is multiplying by $\frac{1}{2}$ , and $x = x^{1/2}$ : $g (x) = \frac{1}{2} x^{2} + 3 x^{1/2} .$ Step 2 — differentiate term by term. $g^{'} (x) = \frac{1}{2} \cdot 2 x + 3 \cdot \frac{1}{2} x^{- 1/2} = x + \frac{3}{2 x} .$

Recall Solution 3.3

Rewrite: $\frac{5}{x ^{3}} = 5 x^{- 3}$ and $\frac{x}{7} = \frac{1}{7} x$ . $y^{'} = 5 (- 3) x^{- 4} - 4 \cdot \frac{1}{2} x^{- 1/2} + \frac{1}{7} = - \frac{15}{x ^{4}} - \frac{2}{x} + \frac{1}{7} .$

Level 4 — Synthesis

Now combine linearity with an extra condition — evaluate a derivative, find a slope, or solve for a point.

Recall Solution 4.1

Step 1 — differentiate (linearity + power rule): $y^{'} = 6 x^{2} - x .$ Step 2 — evaluate at $x = 2$ : $y^{'} (2) = 6 (4) - 2 = 24 - 2 = 22.$ The slope is $22$ .

Recall Solution 4.2

WHY a derivative: "horizontal tangent" means slope $= 0$ , and slope is $f^{'} (x)$ . $f^{'} (x) = 3 x^{2} - 12 = 0 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2.$ Both cases matter: $x = 2$ and $x = - 2$ . Never drop the negative root.

Recall Solution 4.3

Differentiate: $f^{'} (x) = 2 a x + b$ .

$f^{'} (0) = b = 3$ .
$f^{'} (1) = 2 a + b = 7 \Rightarrow 2 a = 4 \Rightarrow a = 2$ . So $a = 2$ , $b = 3$ .

Level 5 — Mastery

Prove and generalize. Here you connect linearity to its deeper home: Linear operators and Definition of the derivative (limit of difference quotient).

Recall Solution 5.1

Let $h (x) = 3 f (x) - 2 g (x)$ . By definition: $h^{'} (x) = lim_{Δ x \to 0} \frac{[ 3 f ( x + Δ x ) - 2 g ( x + Δ x ) ] - [ 3 f ( x ) - 2 g ( x ) ]}{Δ x} .$ Regroup $f$ -terms with $f$ -terms, $g$ with $g$ (addition is commutative): $= lim_{Δ x \to 0} [3 \cdot \frac{f ( x + Δ x ) - f ( x )}{Δ x} - 2 \cdot \frac{g ( x + Δ x ) - g ( x )}{Δ x}] .$ Apply limit laws (sum splits; constants pull out — legal because both limits exist): $= 3 lim_{Δ x \to 0} \frac{f ( x + Δ x ) - f ( x )}{Δ x} - 2 lim_{Δ x \to 0} \frac{g ( x + Δ x ) - g ( x )}{Δ x} = 3 f^{'} (x) - 2 g^{'} (x) . ■$ This is exactly the one-line linearity statement $\frac{d}{d x} [a f + b g] = a f^{'} + b g^{'}$ with $a = 3, b = - 2$ .

Recall Solution 5.2

Take $f (x) = g (x) = x$ . Then $f (x) g (x) = x^{2}$ , so $\frac{d}{d x} [f g] = 2 x$ . But $f^{'} (x) g^{'} (x) = 1 \cdot 1 = 1$ . Since $2 x \neq = 1$ for all $x$ (they agree only at $x = \frac{1}{2}$ ), the naive product-split is false. Products need the Product rule: $(f g)^{'} = f^{'} g + f g^{'} = 1 \cdot x + x \cdot 1 = 2 x$ . ✅

Recall Solution 5.3

General form. Linearity splits the sum and pulls out each coefficient; power rule handles each $x^{k}$ : $P^{'} (x) = \sum_{k = 1}^{n} k a_{k} x^{k - 1} .$ The $k = 0$ term ( $a_{0}$ , the constant) vanishes. Apply to $4 x^{5} - 3 x^{2} + 9$ : $P^{'} (x) = 20 x^{4} - 6 x .$ Evaluate at $x = 1$ : $P^{'} (1) = 20 - 6 = 14.$

Connections

Sum, difference, constant multiple rules — the parent this drill trains.
Definition of the derivative (limit of difference quotient) — used in Exercise 5.1.
Limit laws (sum, scalar, product of limits) — the engine behind that proof.
Power rule — paired with linearity in nearly every problem here.
Product rule — the non-linear sibling, isolated in Exercise 5.2.
Linear operators — the abstract home of Exercise 5.1's structure.
Linearity of integration — the integral twin of the same idea.