IB DP Mathematics: Analysis and Approaches – Limits, derivatives, tangents, rules of differentiation and graph behaviour Study Guide

This study guide teaches the differentiation unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 5.1–5.4 and 5.6–5.8. All of it is SL content, so it is required at both SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

The calculus strand overview gives the big picture of the whole strand. This page goes section by section through differentiation only. Section 5.5 (anti-differentiation and area) is not part of this unit. When you have worked through it, test your recall with the revision notes and the practice questions. The IB DP Maths AA course hub and the printable syllabus checklist show where the unit sits in the course.

What this unit covers

Section	What you must be able to do	SL/HL
5.1	Estimate a limit from a table or graph; read a derivative as a gradient function and a rate of change; use dy/dx, f′(x), dV/dr, ds/dt	SL and HL
5.2	Find intervals where f is increasing (f′(x) > 0) or decreasing (f′(x) < 0)	SL and HL
5.3	Differentiate axⁿ and sums of such terms, n ∈ ℤ	SL and HL
5.4	Find equations of tangents and normals, by hand and with technology	SL and HL
5.6	Differentiate xⁿ (n ∈ ℚ), sin x, cos x, eˣ, ln x; use the chain, product and quotient rules	SL and HL
5.7	Use the second derivative; link the graphs of f, f′ and f″	SL and HL
5.8	Find and test local maxima and minima; solve optimisation problems; find points of inflexion with zero and non-zero gradient	SL and HL

The guide says formal analytic methods of calculating limits are not required at SL. Differentiation from first principles is an HL-only section (5.12) and is not covered here.

5.1 Limits and the meaning of a derivative

A limit is the value a function’s output approaches as the input approaches some value. At SL you estimate it from a table or graph.

The gradient of a curve at a point is the limit of the gradients of chords. Take a chord from the point to a nearby point, then move the nearby point closer and closer.

Worked example: gradient of y = x³ at x = 2

The chord from (2, 8) to (2 + h, (2 + h)³) has gradient ((2 + h)³ − 8)/h.

h	chord gradient
0.1	12.61
0.01	12.0601
0.001	12.006001

As h → 0 the chord gradients approach 12. So the gradient of the curve at x = 2 is 12.

Notation and meaning

The derivative is the gradient function. It gives the gradient of the tangent at any point. It is also the rate of change of one variable with respect to another. The guide lists these forms:

• dy/dx or f′(x) — gradient of y = f(x)
• dV/dr — rate of change of volume V with respect to radius r
• ds/dt — rate of change of displacement s with respect to time t

State units for a rate in context: dV/dr in cm³ per cm.

5.2 Increasing and decreasing functions

• f is increasing on an interval where f′(x) > 0.
• f is decreasing on an interval where f′(x) < 0.
• Where f′(x) = 0 the tangent is horizontal.

Worked example

Find the intervals where f(x) = x³ − 3x² − 9x + 2 is decreasing.

f'(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1)
f'(x) = 0 at x = −1 and x = 3
test x = −2: f'(−2) = 15 > 0
test x = 0:  f'(0)  = −9 < 0
test x = 4:  f'(4)  = 15 > 0

f is decreasing for −1 < x < 3, and increasing for x < −1 and for x > 3.

5.3 Differentiating powers of x

If f(x) = axⁿ then f′(x) = anxⁿ⁻¹, for n ∈ ℤ. Differentiate a sum term by term. A constant differentiates to 0.

Rewrite before you differentiate: 5/x² = 5x⁻², and (x² − 3)/x = x − 3x⁻¹.

Worked example

y = 4x³ − 5/x² + 7 = 4x³ − 5x⁻² + 7
dy/dx = 12x² + 10x⁻³ = 12x² + 10/x³

f(x) = (x² − 3)/x = x − 3x⁻¹
f'(x) = 1 + 3x⁻² = 1 + 3/x²
f'(3) = 1 + 3/9 = 4/3

5.4 Tangents and normals

At the point (x₁, y₁) on y = f(x):

• the tangent has gradient m = f′(x₁), so y − y₁ = m(x − x₁)
• the normal is perpendicular to the tangent, so its gradient is −1/m (when m ≠ 0)

Worked example

Find the tangent and normal to y = x³ − 4x + 1 at x = 2.

y(2) = 8 − 8 + 1 = 1, so the point is (2, 1)
dy/dx = 3x² − 4, so m = 3(4) − 4 = 8
tangent: y − 1 = 8(x − 2)  →  y = 8x − 15
normal gradient = −1/8
normal:  y − 1 = −(1/8)(x − 2)  →  x + 8y − 10 = 0

The guide expects both analytic methods and technology here. On Paper 2 a GDC can give the gradient at a point, but you still write the equation of the line yourself.

5.6 Standard derivatives and the rules

Standard derivatives

f(x)	f′(x)
xⁿ (n ∈ ℚ)	nxⁿ⁻¹
sin x	cos x
cos x	−sin x
eˣ	eˣ
ln x	1/x

The trigonometric results need x in radians. Multiples and sums work as before: d/dx(3 sin x − 2eˣ) = 3 cos x − 2eˣ.

Rational powers: write roots as powers first.

y = x^(3/2) − 6x^(−1/3)
dy/dx = (3/2)x^(1/2) + 2x^(−4/3)

The chain rule

For a composite function y = f(g(x)), let u = g(x). Then

dy/dx = (dy/du) × (du/dx)

Differentiate the outer function, keep the inside unchanged, then multiply by the derivative of the inside.

y = sin(4x + 1)      dy/dx = 4cos(4x + 1)
y = √(2x + 5)        dy/dx = (1/2)(2x + 5)^(−1/2) × 2 = 1/√(2x + 5)
y = ln(x² + 1)       dy/dx = 2x/(x² + 1)

The product rule

If y = uv, where u and v are functions of x, then

dy/dx = u(dv/dx) + v(du/dx)

Worked example: y = x²e³ˣ.

u = x²,   du/dx = 2x
v = e^(3x), dv/dx = 3e^(3x)   (chain rule)
dy/dx = x²(3e^(3x)) + e^(3x)(2x) = xe^(3x)(3x + 2)

The quotient rule

If y = u/v, then

dy/dx = (v(du/dx) − u(dv/dx)) / v²

Worked example: y = (ln x)/x.

u = ln x, du/dx = 1/x
v = x,    dv/dx = 1
dy/dx = (x(1/x) − ln x(1)) / x² = (1 − ln x)/x²

The order in the numerator matters: swapping it changes the sign.

5.7 The second derivative and graph behaviour

The second derivative is the derivative of the derivative. Write it as d²y/dx² or f″(x); you need both notations.

• f″(x) > 0: the gradient is increasing, and the curve is concave-up.
• f″(x) < 0: the gradient is decreasing, and the curve is concave-down.

How the three graphs relate:

On the graph of f	On the graph of f′	On the graph of f″
increasing	above the x-axis	—
decreasing	below the x-axis	—
stationary point	crosses or touches the x-axis	—
concave-up	increasing	above the x-axis
point of inflexion	local maximum or minimum	crosses the x-axis

Worked example

Let f(x) = xe⁻ˣ.

f'(x)  = e^(−x) − xe^(−x) = (1 − x)e^(−x)
f''(x) = −e^(−x) − (1 − x)e^(−x) = (x − 2)e^(−x)

Since e⁻ˣ > 0 for all x, the signs depend only on the brackets. f is increasing for x < 1 and decreasing for x > 1, so there is a maximum at (1, e⁻¹). f is concave-down for x < 2 and concave-up for x > 2, so there is a point of inflexion at (2, 2e⁻²).

5.8 Maxima, minima, inflexion and optimisation

Testing a stationary point

Solve f′(x) = 0, then use either test:

1. First derivative (sign change). f′ changes from + to −: local maximum. From − to +: local minimum. No change of sign: not a maximum or minimum.
2. Second derivative. f″(x) > 0: minimum. f″(x) < 0: maximum. If f″(x) = 0 the test tells you nothing; use the sign-change test.

Points of inflexion

At a point of inflexion f″(x) = 0 and f″ changes sign. The guide stresses that f″(x) = 0 alone is not enough: for y = x⁴, f″(0) = 0 but (0, 0) is a minimum, not an inflexion. An inflexion can have zero gradient (a stationary, or horizontal, point of inflexion) or non-zero gradient.

Worked example: both kinds of inflexion

f(x) = x⁴ − 4x³.

f'(x)  = 4x³ − 12x² = 4x²(x − 3)    zero at x = 0, x = 3
f''(x) = 12x² − 24x = 12x(x − 2)    zero at x = 0, x = 2

• x = 3: f″(3) = 36 > 0, so (3, −27) is a local minimum.
• x = 0: f″(0) = 0, so use signs. f′(−1) = −16 and f′(1) = −8: no sign change, so not a max or min. f″ changes from + (at x = −1, f″ = 36) to − (at x = 1, f″ = −12). So (0, 0) is a point of inflexion with zero gradient.
• x = 2: f″ changes from − to + and f′(2) = −16 ≠ 0. So (2, −16) is a point of inflexion with non-zero gradient.

Optimisation

1. Write the quantity to optimise as a function of one variable (use a constraint to remove the other).
2. Differentiate and solve derivative = 0.
3. Justify max or min (second derivative or sign change).
4. Answer the question asked, with units.

Worked example: a closed cylinder has volume 250π cm³. Find the radius that minimises its surface area S.

V = πr²h = 250π  →  h = 250/r²
S = 2πr² + 2πrh = 2πr² + 500π/r
dS/dr = 4πr − 500π/r² = 0  →  r³ = 125  →  r = 5
d²S/dr² = 4π + 1000π/r³ = 12π > 0 at r = 5, so minimum
h = 250/25 = 10,  S = 50π + 100π = 150π cm²

The minimum surface area is 150π cm², when r = 5 cm and h = 10 cm.

Using your GDC

Paper 1 allows no technology, so every rule in 5.3 and 5.6, and all the algebra of stationary points, must be done by hand. On Paper 2 the guide expects technology for:

• tables of values to estimate limits (5.1)
• the numerical gradient at a point and the equation of a tangent (5.4)
• sketching f, f′ and f″ and locating maxima, minima and zeros of f″ (5.7, 5.8)

Write the expression you are evaluating before the GDC value, so method marks are available.

Common errors

• Differentiating 5/x² as 5/(2x) instead of rewriting as 5x⁻² first.
• Using f′(x₁) as the gradient of the normal instead of −1/f′(x₁).
• Dropping the inner derivative: d/dx(e^(3x)) is 3e^(3x), not e^(3x).
• Reversing the quotient rule numerator.
• Calling a point an inflexion because f″(x) = 0, without checking a sign change.
• Finding x for a stationary point and never giving the y-coordinate or the value asked for.
• Giving an optimisation answer with no justification that it is a maximum or minimum.
• Using degrees with sin x and cos x in calculus.

Go on to the revision notes for condensed recall, then the practice questions. For how calculus links to graph work, see the functions strand overview. The syllabus guide, subject guide and exam preparation page cover the course as a whole.

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020). This page covers syllabus sections 5.1, 5.2, 5.3, 5.4, 5.6, 5.7 and 5.8.