What is the significance of the second derivative in Differentiation 2?

In Differentiation 2, the second derivative is crucial for understanding the concavity of a function and identifying points of inflection. It helps determine whether a function is concave up or concave down at a particular point. In the IB DP Mathematics curriculum, this concept is essential for analyzing the behavior of functions and solving optimization problems.

How do you apply the second derivative test in Differentiation 2?

The second derivative test is used to determine the nature of stationary points. If the second derivative at a stationary point is positive, the function has a local minimum there. If it is negative, the function has a local maximum. If the second derivative is zero, the test is inconclusive, and further analysis is needed. This method is a key part of the IB DP Mathematics syllabus for solving optimization problems.

What are common applications of Differentiation 2 in real-world problems?

Differentiation 2 is widely used in real-world applications such as physics for analyzing motion, in economics for finding cost and revenue optimization, and in engineering for stress and strain analysis. In the IB DP Mathematics course, students learn to apply these concepts to solve practical problems, enhancing their analytical skills.

How does the concept of concavity relate to the second derivative in Differentiation 2?

Concavity is directly related to the second derivative of a function. If the second derivative is positive over an interval, the function is concave up on that interval, resembling a 'U' shape. Conversely, if the second derivative is negative, the function is concave down, resembling an 'n' shape. Understanding concavity is crucial for graph sketching and analysis in the IB DP Mathematics curriculum.

Why is understanding points of inflection important in Differentiation 2?

Points of inflection occur where the concavity of a function changes, which is indicated by the second derivative changing sign. These points are important in understanding the overall shape and behavior of a graph. In the IB DP Mathematics program, identifying points of inflection is essential for comprehensive function analysis and graph interpretation.

Why is "Writing the quotient rule as $\dfrac{u' v + u v'}{v^2}$ (using plus)." a common mistake in Differentiation 2?

Why it happens: Confusing with the product rule. How to avoid it: Quotient: TOP times BOTTOM differentiated SECOND. (u/v)' = u'v - uv'/v^2 — the order matters because of the minus.

Why is "Differentiating $\cos(3x)$ as $-\sin(3x)$ (forgetting the chain factor of 3)." a common mistake in Differentiation 2?

Why it happens: Treating the bracket as a single x. How to avoid it: Apply the chain rule: derivative of OUTER times derivative of INNER. d/dx(3x) = -(3x) · 3.

Why is "Finding stationary points and not classifying them." a common mistake in Differentiation 2?

Why it happens: Treating f' = 0 as the whole answer. How to avoid it: Always classify with f'' or a sign test. State explicitly: 'since f''(x_0) > 0, x_0 is a local minimum'.

Why is "In an optimisation problem, forgetting to state or use the physical domain." a common mistake in Differentiation 2?

Why it happens: Rushing to differentiate. How to avoid it: State the domain BEFORE differentiating (e.g. '0 < x < 15') and reject roots outside it.

DifferentiationIB Maths AA SL

Differentiation 2

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Differentiation 2 — IB Maths AA SL: chain, product, quotient rules, stationary points and optimisation

This note covers the three combining rules of differentiation, the second derivative, the test for stationary points, concavity, and optimisation — the applied side of AA SL calculus.

What you’ll learn

Mapped to the IB DP Maths AA SL subject guide (2021 onwards (applies to 2026 exams)).

AO1 — State the chain, product and quotient rules.
AO2 — Apply combining rules to differentiate composite functions.
AO2 — Find and classify stationary points using the first or second derivative.
AO2 — Solve optimisation problems by building and differentiating a model.
AO3 — Justify whether a stationary point is a maximum, minimum or inflexion.

Chain, product and quotient rules

All three in the formula booklet.

Chain rule. If $y = f(g(x))$ , then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x).$

In Leibniz form: if $y$ is a function of $u$ which is a function of $x$ : $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.$

Worked example. Differentiate $y = (3x^2 + 1)^5$ .

Let $u = 3x^2 + 1$ , $y = u^5$ . $\dfrac{dy}{du} = 5u^4$ , $\dfrac{du}{dx} = 6x$ . So $\frac{dy}{dx} = 5u^4 \cdot 6x = 30x(3x^2 + 1)^4.$

Worked example. Differentiate $y = e^{\sin x}$ .

$\dfrac{dy}{dx} = e^{\sin x} \cdot \cos x$ .

Worked example. Differentiate $y = \ln(2x^2 + 3)$ .

$\dfrac{dy}{dx} = \dfrac{1}{2x^2 + 3} \cdot 4x = \dfrac{4x}{2x^2 + 3}$ .

Product rule. $(uv)' = u'v + uv'$ .

Worked example. $y = x^3 e^x$ . $\dfrac{dy}{dx} = 3x^2 \cdot e^x + x^3 \cdot e^x = e^x(3x^2 + x^3) = x^2 e^x(3 + x)$ .

Quotient rule. $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ .

Worked example. $y = \dfrac{\sin x}{x}$ .

$u = \sin x$ , $v = x$ . $u' = \cos x$ , $v' = 1$ .

$\dfrac{dy}{dx} = \dfrac{x \cos x - \sin x}{x^2}$ .

Identifying the right rule.

Composite ( $f$ of $g$ ) → CHAIN.
Product of two functions → PRODUCT.
Ratio of two functions → QUOTIENT.
Multiple rules can compose: differentiate the outer first, recurse.

Chain rule: derivative of OUTER times derivative of INNER.
Product: $u'v + uv'$ .
Quotient: $\dfrac{u'v - uv'}{v^2}$ — order matters!
Apply rules step by step; don't skip.

Stationary points and classification

Set $f' = 0$ , then check $f''$ or sign change.

A stationary point of $f$ is a point where $f'(x) = 0$ .

Classification with $f''$ (second-derivative test):

$f''(x_0) > 0$ → CONCAVE UP locally → $x_0$ is a LOCAL MINIMUM.
$f''(x_0) < 0$ → CONCAVE DOWN locally → $x_0$ is a LOCAL MAXIMUM.
$f''(x_0) = 0$ → INCONCLUSIVE; use the first-derivative sign test.

First-derivative sign test:

Sign of $f'$ on left of $x_0$	Sign on right	$x_0$ is
$+$	$-$	MAXIMUM
$-$	$+$	MINIMUM
same	same	NOT max/min (inflexion with stationary tangent)

Worked example. Find and classify the stationary points of $f(x) = x^3 - 3x + 2$ .

$f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$ . Set $f'(x) = 0$ : $x = \pm 1$ .

$f''(x) = 6x$ . At $x = 1$ : $f''(1) = 6 > 0 \Rightarrow$ MIN. At $x = -1$ : $f''(-1) = -6 < 0 \Rightarrow$ MAX.

$y$ -coordinates: $f(1) = 0$ , $f(-1) = 4$ .

Stationary points: $(1, 0)$ MIN, $(-1, 4)$ MAX.

Inflexion points. Where $f''$ changes sign. Could be at stationary or non-stationary points.

$f(x) = x^3$ has $f'(0) = 0$ but it's NOT a max/min — it's an inflexion with stationary tangent (sign of $f'$ is positive on both sides).

At each stationary point the tangent is horizontal. $f'' > 0$ identifies a minimum (cup); $f'' < 0$ identifies a maximum (cap).

Stationary points: $f'(x) = 0$ .
$f''$ sign classifies max / min.
First-derivative sign test as backup.
Inflexion = $f''$ changes sign.

Optimisation

Real-world: build the model, differentiate, classify.

Standard process:

Define variables in words.
Write the QUANTITY to optimise in terms of ONE variable (use any constraint).
State the domain (often positivity, physical bounds).
Differentiate; set derivative to zero.
Classify with $f''$ or sign test.
State the answer in CONTEXT with units.

Worked example. A rectangle has perimeter $40$ cm. Find the dimensions that maximise its area.

Let length $= x$ , width $= 20 - x$ (perimeter constraint: $2x + 2w = 40$ ).

Area: $A(x) = x(20 - x) = 20x - x^2$ . Domain: $0 < x < 20$ .

$A'(x) = 20 - 2x = 0 \Rightarrow x = 10$ . $A''(x) = -2 < 0$ → maximum.

Width $= 20 - 10 = 10$ . Maximum area $= 100$ cm² (a square).

Worked example (closed cylinder). A cylinder has volume $1000$ cm³. Find the radius minimising surface area.

Volume: $\pi r^2 h = 1000 \Rightarrow h = \dfrac{1000}{\pi r^2}$ .

Surface area: $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \dfrac{1000}{\pi r^2} = 2\pi r^2 + \dfrac{2000}{r}$ .

$S'(r) = 4\pi r - \dfrac{2000}{r^2} = 0 \Rightarrow r^3 = \dfrac{500}{\pi} \Rightarrow r = \sqrt[3]{\dfrac{500}{\pi}} \approx 5.42$ cm.

$S''(r) = 4\pi + \dfrac{4000}{r^3} > 0$ → minimum confirmed.

Examiner tip. Always state the second-derivative or sign-test confirmation. 'Setting derivative to zero' gives stationary points; the test tells you it's a MINIMUM (not a max or inflexion).

Reduce to ONE variable using the constraint.
Always state the DOMAIN.
Differentiate, set to zero, CLASSIFY.
State the answer in context.

How it’s examined

Paper 1: chain/product/quotient rule fluency, classify simple stationary points. Paper 2: longer optimisation problems with GDC for $f'(x) = 0$ when algebra is messy. Examiner reports stress the importance of stating the classification test and interpreting the answer in context.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-25.

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Step-by-step worked examples — Differentiation 2

Step-by-step solutions to past-paper-style questions on differentiation 2, written exactly the way a tutor would explain them at the board.

1Chain rule with trig
Building confidence• chain
Question
Differentiate $y = \cos(3x^2 + 1)$ .
Step-by-step solution
1. Step 1
  Outer: $\cos u$ where $u = 3x^2 + 1$ . Outer derivative: $-\sin u$ .
2. Step 2
  Inner derivative: $6x$ .
3. Step 3
  Chain: $-\sin(3x^2 + 1) \cdot 6x = -6x \sin(3x^2 + 1)$ .
Answer
$-6x \sin(3x^2 + 1)$ .
2Product rule
Building confidence• product
Question
Differentiate $y = x^2 \ln x$ .
Step-by-step solution
1. Step 1
  $u = x^2$ , $v = \ln x$ . $u' = 2x$ , $v' = 1/x$ .
2. Step 2
  $y' = 2x \ln x + x^2 \cdot \dfrac{1}{x} = 2x \ln x + x$ .
Answer
$y' = 2x\ln x + x = x(2\ln x + 1)$ .
3Quotient rule
Stretch• quotient
Question
Differentiate $y = \dfrac{e^x}{x^2 + 1}$ .
Step-by-step solution
1. Step 1
  $u = e^x$ , $v = x^2 + 1$ . $u' = e^x$ , $v' = 2x$ .
2. Step 2
  $y' = \dfrac{e^x(x^2 + 1) - e^x(2x)}{(x^2 + 1)^2} = \dfrac{e^x(x^2 - 2x + 1)}{(x^2 + 1)^2}$ .
3. Step 3
  Numerator factorises: $x^2 - 2x + 1 = (x-1)^2$ . So $y' = \dfrac{e^x(x-1)^2}{(x^2+1)^2}$ .
Answer
$y' = \dfrac{e^x (x - 1)^2}{(x^2 + 1)^2}$ .
4Stationary points + classification
Stretch• stationary point
Question
Find and classify the stationary points of $f(x) = x^4 - 8x^2 + 5$ .
Step-by-step solution
1. Step 1
  $f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2)$ .
2. Step 2
  Set $f'(x) = 0$ : $x = 0, 2, -2$ .
3. Step 3
  $f''(x) = 12x^2 - 16$ . Evaluate: $f''(0) = -16 < 0$ (MAX); $f''(\pm 2) = 32 > 0$ (MIN).
4. Step 4
  $f(0) = 5$ MAX; $f(\pm 2) = 16 - 32 + 5 = -11$ MIN.
Answer
$(0, 5)$ MAX; $(2, -11)$ and $(-2, -11)$ MIN.
5Optimisation: open box
Stretch• optimisation, AO3
Question
From a square sheet of side 30 cm, identical squares of side $x$ cm are cut from each corner and the sides folded up to form an open box. Find $x$ that maximises the volume.
Step-by-step solution
1. Step 1
  Volume: $V(x) = x(30 - 2x)^2$ , valid for $0 < x < 15$ .
2. Step 2
  $V'(x) = (30 - 2x)^2 + x \cdot 2(30 - 2x)(-2) = (30 - 2x)\big((30 - 2x) - 4x\big) = (30 - 2x)(30 - 6x)$ .
3. Step 3
  $V' = 0$ : $30 - 2x = 0 \Rightarrow x = 15$ (rejected: boundary) or $30 - 6x = 0 \Rightarrow x = 5$ .
4. Step 4
  Test: $V'(4) > 0$ , $V'(6) < 0$ — so $x = 5$ is a MAXIMUM.
5. Step 5
  $V(5) = 5(20)^2 = 2000$ cm³. Maximum volume at $x = 5$ cm.
Answer
$x = 5$ cm; $V_{\max} = 2000$ cm³.

Key Definitions and Keywords — Differentiation 2

Definitions to memorise and the exact keywords mark schemes credit for differentiation 2 answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA SL sitting.

Chain rule
Examiner keyword
$\dfrac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$ .
Product rule
Examiner keyword
$(uv)' = u'v + uv'$ .
Quotient rule
Examiner keyword
$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ .
Stationary point
Examiner keyword
A point where $f'(x) = 0$ . Could be max, min or stationary inflexion.
Second derivative $f''$
The derivative of $f'$ . Sign indicates concavity.

Common Mistakes and Misconceptions — Differentiation 2

The traps other students keep falling into on differentiation 2 questions — taken from recent IB DP Maths AA SL examiner reports and mark schemes — and how to avoid them.

✕Writing the quotient rule as $\dfrac{u' v + u v'}{v^2}$ (using plus).
Why it happens
Confusing with the product rule.
How to avoid it
Quotient: TOP times BOTTOM differentiated SECOND. $(u/v)' = \dfrac{u'v - uv'}{v^2}$ — the order matters because of the minus.
✕Differentiating $\cos(3x)$ as $-\sin(3x)$ (forgetting the chain factor of 3).
Why it happens
Treating the bracket as a single $x$ .
How to avoid it
Apply the chain rule: derivative of OUTER times derivative of INNER. $\dfrac{d}{dx}\cos(3x) = -\sin(3x) \cdot 3$ .
✕Finding stationary points and not classifying them.
Why it happens
Treating $f' = 0$ as the whole answer.
How to avoid it
Always classify with $f''$ or a sign test. State explicitly: 'since $f''(x_0) > 0$ , $x_0$ is a local minimum'.
✕In an optimisation problem, forgetting to state or use the physical domain.
Why it happens
Rushing to differentiate.
How to avoid it
State the domain BEFORE differentiating (e.g. ' $0 < x < 15$ ') and reject roots outside it.

Past paper style quiz

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Differentiation 2 — frequently asked questions

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Differentiation 2

Short Study Notes in the form of Slides

Short Study Notes — Differentiation 2

Detailed Notes

What you’ll learn

How it’s examined

1Chain rule with trig

2Product rule

3Quotient rule

4Stationary points + classification

5Optimisation: open box

Chain rule

Product rule

Quotient rule

Stationary point

Second derivative f′′f''f′′

✕Writing the quotient rule as u′v+uv′v2\dfrac{u' v + u v'}{v^2}v2u′v+uv′​ (using plus).

✕Differentiating cos⁡(3x)\cos(3x)cos(3x) as −sin⁡(3x)-\sin(3x)−sin(3x) (forgetting the chain factor of 3).

✕Finding stationary points and not classifying them.

✕In an optimisation problem, forgetting to state or use the physical domain.

Past paper style quiz

Differentiation 2 — frequently asked questions

Second derivative $f''$

✕Writing the quotient rule as $\dfrac{u' v + u v'}{v^2}$ (using plus).

✕Differentiating $\cos(3x)$ as $-\sin(3x)$ (forgetting the chain factor of 3).