A derivative measures how fast something changes. That one idea answers three kinds of question. How fast is a ripple's area growing? Where is a graph rising and where is it falling? What shape of box, can or window is the best possible?
A derivative tells you how fast one quantity changes when another one changes. That speed of change is called a rate.
If y = f(x), this is the rate of change of y with respect to x: how much y changes per unit change of x. f′(x₀) is that rate at x = x₀. (Derivatives are in Chapter 5.)
When several quantities change with time t, the chain rule links their rates: differentiate with respect to r, then multiply by how fast r changes.
If x = f(t) and y = g(t), divide the two time rates. This needs dx/dt ≠ 0.
In economics, marginal cost and marginal revenue are rates with respect to x, the number of items.
A quantity that is decreasing has a negative rate. In Example 4 the length shrinks at 3 cm/min, so dx/dt = −3.
Rate of change of the area of a circle w.r.t. r at r = 5.
Cube volume grows 9 cm³/s; surface rate at x = 10.
Ripple at 4 cm/s, r = 10.
Rectangle: dx/dt = −3, dy/dt = 2 at x = 10, y = 6.
Marginal cost at x = 3.
Marginal revenue at x = 5.
Where the slope f′ is positive the graph climbs; where it is negative the graph falls. So the sign of f′ tells you up or down.
for any x₁, x₂ in an interval I: moving right never takes you down. If always f(x₁) < f(x₂), f is strictly increasing.
The same with the inequality reversed (strictly decreasing: f(x₁) > f(x₂)). If f(x) = c throughout, f is constant.
(f′ on all of (a, b), as the book prints it.) And f′ = 0 makes f constant. This holds when f is continuous on [a, b] and differentiable on (a, b): the sign of f′ on (a, b) decides f on [a, b].
How to use it
7x − 3 on R.
x³ − 3x² + 4x.
cos x on (0, 2π).
x² − 4x + 6.
4x³ − 6x² − 72x + 30.
sin 3x on [0, π/2].
sin x + cos x on [0, 2π].
The tops of hills and bottoms of valleys sit where the tangent is flat, or where the slope does not exist, like at a sharp corner.
Then f(c) is the maximum value of f on I. With ≤ instead of ≥, it is the minimum value.
c is a local maximum if this holds just in some small interval around c: the top of one hill, not of the whole graph. A local minimum is the same with ≤.
At a local maximum or minimum, either f′(c) = 0 or f is not differentiable at c. Such points are called critical points.
How to use it · first derivative test
x² on R.
|x| on R.
x on (0, 1).
x³ − 3x + 3.
2x³ − 6x² + 6x + 5.
3 + |x|.
The second derivative f″ tells you how the graph bends. At a flat point, a cup shape means a bottom and a cap shape means a top.
c is a local minimum. A positive f″ means the slope is increasing, so the graph curves up like a cup and the flat point must be a bottom.
c is a local maximum. A negative f″ means the slope is decreasing, so the graph curves down like a cap.
The test fails: it does not decide. Go back to the first derivative test.
How to use it · Theorem 4
3x⁴ + 4x³ − 12x² + 12.
2x³ − 6x² + 6x + 5.
On a closed interval the highest and lowest values always exist. List every candidate point, work out f at each, and compare.
Then f always has an absolute maximum and an absolute minimum on [a, b].
If one of them is attained at an interior point c (not an end point), then f′ = 0 there.
On an open interval there may be neither: here there is always a bigger and a smaller value (Example 16).
How to use it · working rule
2x³ − 15x² + 36x + 1 on [1, 5].
12x4/3 − 6x1/3 on [−1, 1].
Helicopter on y = x² + 7, soldier at (3, 7).
To find the best box, fence or can, write the quantity as a function of one variable. Then find its largest or smallest value.
How to use it
Sum 15, least sum of squares.
Shortest distance from (0, c) to y = x².
Poles 16 m, 22 m, 20 m apart.
Trapezium with three sides 10.
Cylinder in a cone, most curved surface.
x = t²(2 − t/3): when does the car stop?
Inverted cone tank, 5 m³/h, depth 4 m.
Man 2 m, lamp 6 m, 5 km/h.
(3/10)x⁴ − (4/5)x³ − 3x² + (36/5)x + 11.
tan⁻¹(sin x + cos x) on (0, π/4).
Disc r = 3.2, dr/dt = 0.05.
Box from a 3 × 8 sheet.
Price 5 − x/100, cost x/5 + 500.
Twelve quick questions to check yourself. Tap an option to check it and see the reasoning.
Nine are every MCQ from Exercises 6.1–6.3 and the miscellaneous exercise; three are quick concept checks.
dy/dx is the rate of change of y with respect to x; with time as a link, dy/dx = (dy/dt)/(dx/dt) when dx/dt ≠ 0.
f′ ≥ 0 on (a, b) ⇒ increasing; f′ ≤ 0 ⇒ decreasing; f′ = 0 ⇒ constant (f continuous on [a, b]).
A critical point is where f′(c) = 0 or f is not differentiable. Local extrema can only happen there.
First derivative test: + to − gives a local max, − to + a local min, no change an inflection.
Second derivative test: f′(c) = 0 with f″(c) < 0 gives a max, f″(c) > 0 a min; f″(c) = 0 means the test fails.
Absolute extrema on [a, b]: evaluate f at the critical points and the end points; take the largest and the smallest.
Before calculus had a name, Fermat found maxima and minima by noticing that near a highest point a small change hardly changes the value: the idea behind f′ = 0.
His first calculus paper was titled, roughly, "a new method for maxima and minima, and for tangents".
The first calculus textbook explained how to find the highest and lowest points of curves.
He stated the Mean Value Theorem, the fact that turns "f′ > 0" into "f is increasing".
Each example is a small animated story. Circles, cubes, tanks and shadows grow in front of you while the rates are linked by the chain rule. Sign charts fill in piece by piece and then colour the graph green where it rises and red where it falls. In optimisation problems the shape itself changes as you move the design variable, and a point slides along the graph of the quantity to its best value. Press Next step (or Play). Under the controls, each step is explained in detail, with links to the exact basics it uses; after reading a basic, press ← Back to return to the same step.
A derivative is a rate. When quantities change with time, the chain rule links their rates. Marginal cost and revenue are rates too.
f′ > 0 means rising, f′ < 0 falling. Find where f′ = 0, split the line into pieces and test one point in each.
Local maxima and minima sit at critical points. The first derivative test watches the sign of f′; the second derivative test looks at f″.
Write the quantity as a function of one variable, set its derivative to 0, and check it really is a maximum or a minimum.
Compare f at the critical points and at the two end points.
Velocity, filling a tank, a shadow, sign charts, a heated disc, the biggest box and the best profit.
A spreading ripple, the cost of baking one more cake, the top of a hiking trail, the biggest box from a sheet of card, the most economical drink can and a lengthening shadow: each is a rate of change or a search for the best value.
A stone drops into a pond and the ripple's radius grows at 4 cm/s. The area grows at dA/dt = 2πr · 4: the bigger the ring, the more water surface it sweeps each second.
A bakery's daily cost for x cakes is C(x) = 0.01x³ − 0.9x² + 40x + 500 rupees. The marginal cost C′(x) is the slope of the cost curve: very nearly the extra cost of the next cake. It falls at first (bulk buying) and then rises (overtime, a crowded oven).
The height of a hiking trail against distance walked. Uphill stretches (green) have a positive slope, downhill ones (red) a negative slope. The local summits and valleys are where the slope is 0; the very highest point of the whole trail may be a summit or one of the two ends.
A4 card is 21 cm by 29.7 cm. Cut a square of side x from each corner and fold. Too small a cut gives a flat tray; too large a cut, a narrow tower. The derivative finds the best cut.
For a fixed volume V = πr²h = 330 cm³, the metal needed is S = 2πr² + 2V/r. The least metal is used when h = 2r (Exercise 6.3 Q21). Real cans are taller and thinner than this, partly because their ends are made of thicker metal and a thinner can is easier to hold.
A 1.5 m child walks at 1.2 m/s away from a 4.5 m lamp. Similar triangles give s/(l + s) = 1.5/4.5, so l = 2s. The shadow grows at ds/dt = 0.6 m/s, and its tip moves at 1.8 m/s, however far away the child is (Example 32 has the same shape).
Decide first, then tap each card.
Every question that asks you to find, show or prove something has an animated walkthrough here; the nine multiple-choice questions are in the Lesson’s self-check. Pick a question from the chips above each player. Press Play to watch it solved, or step through with Next. Each step has a short "why" under it and links to the basics it needs.
Write the quantity in terms of the changing length, differentiate with respect to time (chain rule), then substitute the given instant. (Q17–Q18 are in the quiz.)
Find f′, find where it is 0 (or undefined), and test the sign on each piece. (Q12, Q13, Q19 are in the quiz.)
Local extrema with the first or second derivative test; absolute extrema on closed intervals by comparing values; word problems by building a function of one variable. (Q27–Q29 are in the quiz.)
Rates, intervals, extrema and optimisation, including a proof with the Mean Value Theorem. (Q16 is in the quiz.)
This chapter uses the derivative from Chapter 5 as a tool. To use it you need a few old ideas, each explained here from its roots: what a rate is, how rates chain together, the area and volume formulas, intervals, signs of products, solving equations, where sin and cos are positive, and how to turn a word problem into a function. Every example and exercise links here exactly where you need it, and a button takes you straight back to the problem.
A car covers s(t) metres in t seconds. Between t and t + h its average speed is [s(t + h) − s(t)]/h. Shrink h and the average settles on the speed at the instant t: that limit is the derivative ds/dt. The same idea works for any pair of quantities: dy/dx is "how much y changes per unit change of x", right now.
Units: a rate has the units of "top per bottom": cm² per second, rupees per item. A negative rate means the quantity is decreasing.
Also for fractional and negative n: (√x)′ = 1/(2√x), (1/x)′ = −1/x². A constant has derivative 0.
(uv)′ = u′v + uv′, (u/v)′ = (u′v − uv′)/v², and d/dx f(g(x)) = f′(g(x)) · g′(x).
(sin x)′ = cos x, (cos x)′ = −sin x, (tan x)′ = sec²x, (eˣ)′ = eˣ, (log x)′ = 1/x, (tan⁻¹x)′ = 1/(1 + x²).
Each rule is proved in the Chapter 5 page. The second derivative f″ is the derivative of f′: the rate at which the slope itself changes.
Suppose each extra centimetre of radius adds 2πr cm² of area, and the radius grows 4 cm every second. Then every second the area grows by 2πr × 4 cm². That is the chain rule: dA/dt = (dA/dr)(dr/dt). We differentiate the formula with respect to the length, then multiply by how fast the length changes.
Always ask two questions: "how does the quantity change when the length changes?" and "how fast does the length change?". Multiply the answers.
Circle: area πr², circumference 2πr. Rectangle: area xy, perimeter 2(x + y). Trapezium: ½(sum of parallel sides) × height.
Cube: volume x³, surface 6x². Open box from a sheet: (length − 2x)(width − 2x) × x.
Sphere: (4/3)πr³, surface 4πr². Cylinder: πr²h, curved surface 2πrh. Cone: (1/3)πr²h, curved surface πrl with slant l = √(r² + h²).
A cone of water inside a conical tank, a man and his shadow under a lamp: the small triangle has the same shape as the big one, so their sides are in the same ratio. That ratio lets us write one length in terms of another (r = h/2, l = 2s) and keep a single variable.
A round bracket leaves the end out; a square bracket includes it.
| Written | Means | |
|---|---|---|
| (a, b) | a < x < b | open: ends left out |
| [a, b] | a ≤ x ≤ b | closed: ends included |
| [a, b) | a ≤ x < b | half-open |
| (a, b] | a < x ≤ b | half-open |
∞ is not a number you can reach, so it always gets a round bracket: (2, ∞). A ∪ B means "in A or in B".
The difference matters: on a closed interval a continuous function always has a greatest and a least value; on an open one it may not (f(x) = x on (0, 1) has neither).
(x + 2)(x − 3) is 0 only at x = −2 and x = 3. Between these "cut points" neither factor can change sign, so the product keeps one sign on each piece. Test one number per piece, or count the negative factors. A squared factor such as (x − 1)² is never negative, so it never changes the sign.
To solve a quadratic, factorise it or use the formula. The discriminant tells you how many roots to expect.
For ax² + bx + c, look for two numbers with product ac and sum b. If none come to mind, use the formula.
Positive: two roots. Zero: one repeated root (a square). Negative: no real roots, and the quadratic keeps one sign for all x (the sign of a).
Take out a common factor first; what is left is a quadratic.
The point at angle x on the unit circle is (cos x, sin x). So where the point is decides the signs.
| Function | Positive in |
|---|---|
| cos x | the right half: first and fourth quadrants |
| sin x | the top half: first and second quadrants |
| tan x = sin xcos x | first and third quadrants |
Equations such as sin x = cos x (that is, tan x = 1) have one solution in each half-turn: π/4 and 5π/4 in [0, 2π].
(1) Name what is to be made largest or smallest. (2) Draw a picture and name the lengths. (3) Use the condition you are given (a fixed perimeter, a fixed volume, a sum of 15) to get rid of every variable but one. (4) Note the allowed range of that variable. Only then differentiate.
Leave your email and we will write when a new chapter or a new set of worked solutions is ready. That is all we use it for, and you can ask to be taken off the list at any time.
You are on the list. We will email you when new chapters are added, and for nothing else.