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IB DP Maths AI · Topic 5 Calculus · SL 5.1–5.4, 5.6, 5.7 · AHL 5.9, 5.10 · ~8 min read

Differentiation and optimisation

The derivative as a rate of change and a gradient, differentiating axⁿ, increasing and decreasing functions, tangents and normals, stationary points, and optimisation in context, plus for HL: derivatives of sin, cos, tan, eˣ and ln x, the chain, product and quotient rules, related rates and the second derivative.

Derivative as rate of changeLimits & gradientsDifferentiating axⁿIncreasing & decreasingTangents & normalsStationary pointsLocal max & minOptimisation
RSby Riyaz Sir
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No tag = SL and HLHL ONLY HL students onlyFORMULA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • The derivative f′(x) = dydx is the instantaneous rate of change of y with x, and the gradient of the tangent to the curve.
  • Power rule: if f(x) = axn, then f′(x) = anxn − 1. Differentiate each term separately.
  • f′(x) > 0 means increasing; f′(x) < 0 means decreasing; f′(x) = 0 at stationary points (local maxima and minima).
  • Optimisation: write the quantity as a function of one variable, set its derivative to 0 (or use the GDC's maximum/minimum), then check it is a maximum or minimum and answer in context.
  • HL: derivatives of sin x, cos x, tan x, ex and ln x; the chain, product and quotient rules; related rates; and the second derivative for concavity and classifying stationary points.

By the end you can

  • Interpret the derivative as a rate of change and as a gradient, in context.
  • Differentiate functions made of terms axⁿ (n an integer).
  • Find where functions are increasing or decreasing.
  • Find equations of tangents and normals.
  • Find and classify stationary points, and solve optimisation problems in context.
  • Differentiate trig, exponential and logarithmic functions, and use the chain, product and quotient rules. HL
  • Solve related rates problems. HL
  • Use the second derivative to classify stationary points and describe concavity. HL

1The derivative

  • As a limit: the gradient of the chord from x to x + h tends to the gradient of the tangent as h → 0. In AI you meet this as an idea, often from a GDC table of chord gradients.
  • In context: if V(t) is the volume of water in a tank (litres) at time t (minutes), V′(t) is the rate of flow in litres per minute. A negative value means the tank is emptying.
f(x) = axn ⇒ f′(x) = anxn − 1 FORMULA BOOKLET
n is an integer at SL
Worked example

Differentiate f(x) = 4x3 − 5x2 + 2x − 7.

  1. rewrite 2x as 2x−1 M1
  2. f′(x) = 12x2 − 10x − 2x−2 = 12x2 − 10x − 2x2 A1A1
Common mistakeDifferentiating 2x as "2". Rewrite fractions with negative powers first: 2x = 2x−1 ⇒ −2x−2.

2Tangents and normals

xy123456246(4, 4)tangent: y = 5x − 16normal: y = −0.2x + 4.8y = x² − 3x
The tangent has the gradient of the curve at the point; the normal is perpendicular to it.
Worked example

Find the equations of the tangent and normal to y = x2 − 3x at x = 4.

  1. y(4) = 4; dydx = 2x − 3 = 5 at x = 4 A1A1
  2. tangent: y − 4 = 5(x − 4), y = 5x − 16 A1
  3. normal gradient −15: y = −0.2x + 4.8 A1
GDC (calculator) methodThe GDC's dy/dx (or nDeriv) tool gives the gradient at a point, and many GDCs draw the tangent line for you. Still write the gradient and the point you used.

3Stationary points

xy0.511.522.53246max (1, 6)min (2, 5)sign of f′:+ (rising)− (falling)+ (rising)f(x) = 2x³ − 9x² + 12x + 1: f′(x) = 0 where the tangent is horizontal.
The sign of f′(x) changes from + to − at a maximum and from − to + at a minimum.
Worked example

f(x) = 2x3 − 9x2 + 12x + 1. Find and classify the stationary points, and state where f is decreasing.

  1. f′(x) = 6x2 − 18x + 12 = 6(x − 1)(x − 2) A1
  2. f′(x) = 0 at x = 1 and 2, giving (1, 6) and (2, 5) M1A1
  3. f′ changes + to − at x = 1 (maximum) and − to + at x = 2 (minimum) R1
  4. decreasing for 1 < x < 2 A1
  • Local vs global: a local maximum is the highest point nearby. On a closed interval, also check the end points: the global maximum might be at an end.

4Optimisation

Worked example

A farmer has 100 m of fencing to make a rectangular pen against a long wall (no fence needed along the wall). Find the dimensions for the largest area.

  1. two sides y and one side 100 − 2y, so A = y(100 − 2y) = 100y − 2y2 M1A1
  2. A′(y) = 100 − 4y = 0 ⇒ y = 25 M1
  3. 25 m by 50 m, area 1250 m2 A1
r (cm)S (cm²)12345678100200300400min S = 264 cm² at r = 3.74V = πr²h = 330 ⇒ h = 330/(πr²)S = 2πr² + 660/rS′(r) = 0 at r = 3.74
Minimum surface area for a fixed volume: the GDC's minimum tool or S′(r) = 0 gives the same answer.
Worked example

A closed cylindrical can must hold 330 cm3. Find the radius that uses the least metal, and that surface area.

  1. h = 330πr2, so S = 2πr2 + 2πrh = 2πr2 + 660r M1A1
  2. S′(r) = 4πr − 660r2 = 0 ⇒ r = 3.74 cm M1A1
  3. Smin = 264 cm2 (h = 7.49 cm) A1
  • Profit models: P(x) = −2x2 + 120x − 1000 has P′(x) = −4x + 120 = 0 at x = 30, so the maximum profit is P(30) = 800.
Exam tipAlways justify that your answer is a maximum (or minimum): a sign change in the derivative, the shape of the graph, or the second derivative (HL). Then answer the question asked, with units.

5More derivatives and rules HL

f(x)f′(x)
xn (any real n)nxn − 1
sin xcos x
cos x−sin x
tan x1cos2x
exex
ln x1x
chain: dydx = dydu × dudx; product: ddx(uv) = udvdx + vdudx; quotient: ddx(uv) = v du/dx − u dv/dxv2 FORMULA BOOKLET
the three rules
Worked example

Differentiate y = x2e3x, y = ln(x2 + 1) and y = sin 2xx.

  1. product: 2xe3x + 3x2e3x M1A1
  2. chain: 2xx2 + 1 A1
  3. quotient: 2x cos 2x − sin 2xx2 M1A1
  • Related rates: link the rates with the chain rule. A balloon's radius grows at 0.5 cm s−1. Since V = 43πr3, dVdt = 4πr2drdt; at r = 10 cm, dVdt = 4π(100)(0.5) = 628 cm3 s−1.

6The second derivative HL

  • f″(x) is the rate of change of the gradient. f″(x) > 0: concave up (gradient increasing); f″(x) < 0: concave down. A point of inflection is where the concavity changes.
  • Second derivative test: at a stationary point, f″ < 0 means a maximum, f″ > 0 a minimum. For f above, f″(x) = 12x − 18: f″(1) = −6 (max), f″(2) = 6 (min); the point of inflection is at x = 1.5.
Linking questionIntegration reverses differentiation (Integration, areas and kinematics). Optimisation uses the models of Modelling with functions, and HL differential equations are built from derivatives (Differential equations and phase portraits). For more algebraic calculus, see Maths AA Differentiation.
Idea for your IA / explorationOptimise something real: the dimensions of a cereal box or drinks can that use the least card or metal for a given volume, and compare your optimum with the real product. Explain why the manufacturer might choose differently.

Test yourself — cover the answers first

Quick check

  1. Differentiate 5x4.
  2. Find the gradient of y = x³ at x = 2.
  3. f′(3) = 0 and f′ changes from − to + at 3. What is at x = 3?
  4. What is the gradient of the normal if the tangent's gradient is 4?
  5. HL Differentiate e5x.
  6. HL f″(a) < 0 at a stationary point. Max or min?

Exam-style

  1. The height of a plant is h(t) = 0.2t3 − 3t2 + 15t + 2 cm, for 0 ≤ t ≤ 8 weeks. Find h′(t), the rate of growth at t = 2, and the time when the growth rate is smallest.
  2. An open box has a square base of side x cm and volume 500 cm³. Show that the surface area is A = x² + 2000x, and find the minimum area.
Show answers
  1. 20x³
  2. 12
  3. a local minimum
  4. −¼
  5. 5e5x
  6. maximum
  7. h′(t) = 0.6t2 − 6t + 15
    h′(2) = 2.4 − 12 + 15 = 5.4 cm per week
    h′ is smallest at its vertex t = 61.2 = 5 weeks (h′(5) = 0)
  8. height = 500x2, so A = x2 + 4x × 500x2 = x2 + 2000x AG
    A′ = 2x − 2000x2 = 0 ⇒ x = 10
    Amin = 300 cm²
Practise this chapter
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.