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Upper secondary · further topics

Derivatives – rates of change and differentiation

Derivatives help you describe how quickly something changes. Start with rates of change and continue with differentiation rules.

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What does the derivative at a point mean?

The derivative f′(a) describes the function's rate of change at x = a. On a graph, it is the slope of the tangent at the point (a, f(a)).

f′(3) = 5 therefore means that the tangent's slope is 5 at x = 3. It does not say that f(3) = 5. f(3) is the function's value, whereas f′(3) describes how quickly that value changes there.

A positive derivative means the function is increasing at the point, and a negative derivative means it is decreasing. The derivative can have different values at different x values.

5.1Growth and rates of change

Think speed! Rate of change is exactly the same idea as speed in physics: how much something changes per unit of time. When we measure distance, we call it speed. When it is something else (kronor, height, quantity), we call it a rate of change — but the mathematics is the same.

Growth is how much a quantity changes. Rate of change is how quickly it changes per unit of time.

time y t₁ t₂ Δt Δy

Growth between two points in time:

Δy = y(t₂) − y(t₁)

Rate of change = growth per unit of time:

Rate of change = Δy / Δt

Think of a car's speed: y is distance, t is time. Δy/Δt is average speed.

Example

A plant is 5 cm tall at the start (t = 0) and 17 cm tall after 4 weeks.

Growth: Δy = 17 − 5 = 12 cm
Rate of change: Δy/Δt = 12 cm / 4 weeks = 3 cm/week
Interpretation: on average, the plant grew 3 cm per week.
Units matter! A rate of change always has a unit — SEK/day, m/s, cm/week. Include the unit in your answer.

5.2Average rate of change

The average rate of change of a function f between two x values is the slope of the secant through the two points.

Average rate of change = (f(b) − f(a)) / (b − a)
(a, f(a)) (b, f(b)) secant

The secant is the line that cuts through the curve at the two points. Slope of the secant = average rate of change.

Other common notation:

Δf/Δx = (f(x + h) − f(x)) / h

Example

For f(x) = x², find the average rate of change from x = 1 to x = 4.

Calculate f(1) and f(4). f(1) = 1, f(4) = 16
Substitute into the formula. (f(4) − f(1)) / (4 − 1) = (16 − 1) / 3
Calculate. 15 / 3 = 5
Answer: the average rate of change is 5.

5.7Differentiating polynomial functions

For polynomials, you only need these rules:

Constant rule
(c)' = 0
E.g. (7)' = 0
The x rule
(x)' = 1
E.g. y = x ⇒ y' = 1
Power rule
(xⁿ)' = n · xⁿ⁻¹
E.g. (x⁵)' = 5x⁴
Constant multiple
(c · f)' = c · f'
E.g. (4x³)' = 4·3x² = 12x²
Sum rule
(f + g)' = f' + g'
Differentiate term by term
Difference rule
(f − g)' = f' − g'
Also term by term

How to do it in practice

Differentiate each term separately using the power rule:

Bring the exponent down as a factor.
Decrease the exponent by 1.
The constant term disappears.

Example

f(x)f'(x)
50
x1
x²2x
x³3x²
4x³12x²
x³ − 3x + 53x² − 3
2x⁴ − 7x² + x − 18x³ − 14x + 1
(1/2)x² + 3xx + 3

Step by step: differentiate f(x) = 2x⁴ − 7x² + x − 1

Differentiate each term. (2x⁴)' = 2 · 4x³ = 8x³ (−7x²)' = −7 · 2x = −14x (x)' = 1 (−1)' = 0
Combine. f'(x) = 8x³ − 14x + 1
Common mistake: forgetting to multiply by the exponent. (x⁵)' is not x⁴ — it is 5x⁴.
Here we apply the rules to polynomials.

Functions such as sin x, cos x, ex, ln x, quotients (f/g) and composite functions (e.g. (2x+1)5) have their own differentiation rules. These are covered in more advanced mathematics. Here you practise the polynomial rules above. GeoGebra can help with other functions.

5.6Practical uses of differentiation

The derivative is the key to finding maximum and minimum points, the steepest point, and how the function behaves.

Find maximum and minimum points

Differentiate the function. f(x) → f'(x)
Set f'(x) = 0 and solve. These x values are candidates for a maximum/minimum.
Create a sign table for f'(x).
  • f' changes from + to −: maximum point
  • f' changes from − to +: minimum point
  • f' keeps the same sign: stationary point of inflection
Substitute x back into f to find the y coordinate.

Example: f(x) = x³ − 3x

Differentiate. f'(x) = 3x² − 3
Set f'(x) = 0. 3x² − 3 = 0 ⇒ x² = 1 ⇒ x = ±1
Sign table for f'(x) = 3(x − 1)(x + 1):
Intervalx < −1−1 < x < 1x > 1
f'(x)+−+
f(x)increasesdecreasesincreases
Interpret: f' changes from + to − at x = −1 → maximum point. f' changes from − to + at x = 1 → minimum point.
Y coordinates: f(−1) = (−1)³ − 3(−1) = −1 + 3 = 2 f(1) = 1 − 3 = −2
Answer: Maximum point (−1, 2), minimum point (1, −2).

Equation of the tangent

The tangent to the graph at the point (a, f(a)) has slope f'(a) and passes through that point. Using point–slope form:

y − f(a) = f'(a) · (x − a)

Optimisation

Many word problems ask: «What is the greatest/smallest …?» This is optimisation. Strategy:

  1. Write an expression for what you want to maximise/minimise as a function of one variable.
  2. Write down the domain — which values are physically possible?
  3. Differentiate.
  4. Set the derivative = 0 and find the candidates.
  5. Check whether it is a maximum or minimum point (sign table).
  6. Also check the endpoints of the domain — the maximum/minimum may be there!
Do not forget the endpoints! The maximum may occur where the derivative = 0, or at an endpoint of the domain.

Example 1: maximise area (a farmer with 100 m of fencing)

A farmer has 100 m of fencing and wants to enclose a rectangular field. How large can it be?

Variables: length l, width b. Perimeter: 2l + 2b = 100, so b = 50 − l.
Domain: 0 < l < 50 (both sides must be positive).
Area as a function of l: A(l) = l·(50 − l) = 50l − l²
Differentiate: A'(l) = 50 − 2l
Set A' = 0: 50 − 2l = 0 ⇒ l = 25
Check: A'(20) = 10 > 0, A'(30) = −10 < 0. Changes from + to − → maximum point.
Check the endpoints: at l = 0 and l = 50, the area is 0. The maximum point inside the interval wins.
Answer: a 25 × 25 m square gives the largest area = 625 m².

Example 2: minimise material use

An open rectangular tank with a square base must hold 32 litres. What is the smallest possible surface area? (Base + 4 sides.)

Variables: base side x dm, height h dm. Volume: x²·h = 32, so h = 32/x².
Surface area (one base + four sides): O(x) = x² + 4xh = x² + 128/x
Domain: x > 0.
Differentiate (here we must use d/dx(1/x) = −1/x²): O'(x) = 2x − 128/x²
Set O' = 0: 2x = 128/x² ⇒ 2x³ = 128 ⇒ x³ = 64 ⇒ x = 4
Check the minimum point: O'(2) = 4 − 32 = −28 < 0, O'(6) = 12 − 128/36 ≈ 8.4 > 0. Changes from − to + → minimum point.
Answer: the base is 4 × 4 dm, the height h = 32/16 = 2 dm. Minimum surface area = 16 + 32 = 48 dm².

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