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IB Diploma Mathematics: Analysis & Approaches HL · Calculus
Mini-Lesson

Calculus

This HL mini-lesson sharpens differentiation and integration: the chain, product and quotient rules, the second derivative and points of inflection, optimisation, areas between curves, and integration by substitution.

Answer as you go and collect ⭐ stars. Press Start when ready.

Chain rule

The chain rule

To differentiate a function of a function, multiply the outer derivative by the inner derivative:

d/dx f(g(x)) = f′(g(x))·g′(x)e.g. d/dx (2x+1)³ = 3(2x+1)²·2
Quick check

Chain rule

?The derivative of f(g(x)) is:
Product rule

The product rule

For a product of two functions u and v:

(uv)′ = u′v + uv′e.g. d/dx (x²(x+1)) = 2x(x+1) + x² = 3x² + 2x
Calculate

Chain rule

✎Differentiate y = (2x + 1)³ and evaluate dy/dx at x = 1.
Hint: dy/dx = 3(2x+1)²·2; at x=1, 2x+1 = 3.
Quotient rule

The quotient rule

For a quotient u/v:

(u/v)′ = (u′v − uv′)/v²e.g. d/dx (x/(x+1)) = 1/(x+1)²
Quick check

Product rule

?For y = u·v, dy/dx equals:
Calculate

Product rule

✎Differentiate y = x²(x + 1) and evaluate dy/dx at x = 2.
Hint: dy/dx = 2x(x+1) + x² = 3x² + 2x; put x = 2.
Standard derivatives

Derivatives of standard functions

Combine the rules with the standard derivatives:

d/dx: sinx→cosx, cosx→−sinx, eˣ→eˣ, lnx→1/x, tanx→sec²x
Sort it

Which rule differentiates it?

Tap an expression, then the rule you would use.

⛓️ Chain rule

✖️ Product rule

➗ Quotient rule

Second derivative

Concavity & inflection

f″(x) measures concavity: f″ > 0 concave up, f″ < 0 concave down. A point of inflection is where concavity changes (f″ = 0 and changes sign).

Example: y = x³ has y″ = 6x, which is zero and changes sign at x = 0 — a point of inflection.

Calculate

Quotient rule

✎Differentiate y = x/(x + 1) and evaluate dy/dx at x = 1.
Hint: dy/dx = 1/(x+1)²; at x = 1 that is 1/4.
Quick check

Inflection

?A point of inflection occurs where:
Optimisation

Optimisation at HL

Model the quantity, differentiate, solve f′(x) = 0, and use f″ (or a sign test) to confirm a maximum or minimum. Check the domain and endpoints.

Calculate

Inflection

✎Find the x-coordinate of the point of inflection of y = x³.
Hint: y″ = 6x; set 6x = 0.
Match it

Match each rule to its formula

Tap a statement on the left, then its matching partner on the right.

Rule
Formula
Areas

Areas between curves & substitution

The area between two curves is the integral of (top − bottom):

Area = ∫ab (top − bottom) dx∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6

Substitution: for ∫ f(g(x))g′(x) dx, let u = g(x) to simplify.

Calculate

Area between curves

✎Find the area enclosed between y = x and y = x² from x = 0 to x = 1, to 3 d.p.
Hint: ∫₀¹ (x − x²) dx = 1/2 − 1/3.
Quick check

Area between curves

?The area between two curves is found by integrating:
Strategy

Picking the technique

Spot the structure: composition → chain rule; product → product rule; quotient → quotient rule. For area between curves, always integrate top minus bottom over the correct limits.

Exam habit: simplify derivatives before substituting values, and sketch to decide which curve is on top.

Recap

The big ideas to know

Chain: d/dx f(g(x)) = f′(g(x))g′(x)

Product: (uv)′ = u′v + uv′

Quotient: (u/v)′ = (u′v − uv′)/v²

Second derivative: concavity; inflection where f″ = 0 and changes sign

Area: ∫(top − bottom) dx; substitution for f(g(x))g′(x)

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