What is the basic concept of differentiation in IB DP Mathematics?

Differentiation in IB DP Mathematics is a fundamental concept that involves finding the derivative of a function. It measures how a function changes as its input changes, essentially providing the rate of change or the slope of the function at any given point. This is crucial for understanding the behavior of functions and solving problems related to rates of change.

How do you find the derivative of a polynomial function in Differentiation 1?

To find the derivative of a polynomial function in Differentiation 1, apply the power rule. For a term ax^n, the derivative is n*ax^(n-1). For example, the derivative of 3x^2 is 6x. This rule is applied to each term of the polynomial separately, and the results are summed to get the derivative of the entire function.

Why is understanding the derivative important in the IB DP curriculum?

Understanding the derivative is crucial in the IB DP curriculum because it allows students to analyze and interpret the behavior of functions. It is used to find local maxima and minima, solve optimization problems, and understand motion and change in various contexts. Mastery of differentiation is essential for success in higher-level mathematics and related fields.

What are some common applications of differentiation in real-world scenarios?

Differentiation has numerous real-world applications, including calculating velocity and acceleration in physics, optimizing profit and cost functions in economics, and determining the rate of reaction in chemistry. In the IB DP Mathematics curriculum, students explore these applications to understand how mathematical concepts are used to solve practical problems.

What are the common mistakes to avoid when learning Differentiation 1?

Common mistakes in learning Differentiation 1 include misapplying the power rule, neglecting to simplify expressions, and forgetting to apply the chain rule when necessary. Students should also be careful with algebraic manipulations and ensure they understand the underlying concepts rather than just memorizing formulas. Practice and careful review of each step can help avoid these errors.

Why is "$d/dx(\sin 3x) = \cos 3x$." a common mistake in Differentiation 1?

Why it happens: Forgetting the chain rule. How to avoid it: Inner derivative is 3: answer is 3 3x.

Why is "Writing $(u/v)' = (uv' - u'v)/v^2$." a common mistake in Differentiation 1?

Why it happens: Sign swap. How to avoid it: Mnemonic: 'low d high minus high d low' — (u'v - uv')/v^2.

Why is "Using the tangent gradient as the normal gradient." a common mistake in Differentiation 1?

Why it happens: Rushing. How to avoid it: Normal gradient = -1/tangent gradient (perpendicular).

IB DP Mathematics (AA)

Differentiation 1

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Short Notes - Differentiation 1

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Detailed Study Notes

Detailed notes on Differentiation for IB DP Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Differentiation 1 — IB Maths AA HL: limits, derivative from first principles, standard derivatives, chain/product/quotient rules

AA HL deepens differentiation with limits, derivative from first principles, an extended derivative bank (trigonometric, exponential, logarithmic, inverse trig) and the three main rules — preparing for implicit and parametric derivatives in Differentiation 2.

At a glance

DERIVATIVE FROM FIRST PRINCIPLES: $f'(x) = \lim_{h \to 0} [f(x+h) - f(x)]/h$ .
POWER: $\dfrac{d}{dx}(x^n) = nx^{n-1}$ .
TRIG: $\dfrac{d}{dx}(\sin x) = \cos x$ ; $(\cos x)' = -\sin x$ ; $(\tan x)' = \sec^2 x$ .
EXP/LOG: $\dfrac{d}{dx}(e^x) = e^x$ ; $(\ln x)' = 1/x$ .
CHAIN: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ .
PRODUCT: $(uv)' = u'v + uv'$ .
QUOTIENT: $(u/v)' = (u'v - uv')/v^2$ .

What you’ll learn

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

AO1 — Compute limits and recognise indeterminate forms.
AO1 — State and apply the derivative from first principles.
AO1 — List standard derivatives.
AO2 — Apply chain, product and quotient rules to composite functions.
AO2 — Find tangent and normal equations.

Limits and derivative from first principles

Definition and direct evaluation.

Limit. $\lim_{x \to a} f(x) = L$ means $f$ approaches $L$ as $x$ approaches $a$ from either side.

Derivative definition. For $f$ differentiable at $x$ :

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.$

Worked example. Find $f'(x)$ from first principles for $f(x) = x^2$ .

$f'(x) = \lim_{h \to 0} \frac{(x + h)^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x.$

Worked example. Find $f'(x)$ from first principles for $f(x) = 1/x$ .

$\frac{f(x + h) - f(x)}{h} = \frac{1/(x+h) - 1/x}{h} = \frac{-h/[x(x+h)]}{h} = \frac{-1}{x(x+h)}.$

Taking $h \to 0$ : $f'(x) = -1/x^2$ . ✓ matches power rule.

First principles tests rigorous algebra.
Always show the limit explicitly.

Chain, product and quotient rules

Composite function differentiation.

Standard derivatives (memorise).

$f(x)$	$f'(x)$
$x^n$	$nx^{n-1}$
$e^x$	$e^x$
$\ln x$	$1/x$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\arcsin x$	$1/\sqrt{1 - x^2}$
$\arccos x$	$-1/\sqrt{1 - x^2}$
$\arctan x$	$1/(1 + x^2)$

Chain rule. If $y = f(g(x))$ , then $y' = f'(g(x)) \cdot g'(x)$ .

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

Quotient rule. $(u/v)' = (u'v - uv')/v^2$ .

Worked example (chain). Differentiate $y = \sin(3x^2 + 1)$ .

Inner $u = 3x^2 + 1$ , $du/dx = 6x$ . $dy/du = \cos u$ . So $dy/dx = 6x\cos(3x^2 + 1)$ .

Worked example (product). $y = x^2 e^x$ .

$y' = 2x \cdot e^x + x^2 \cdot e^x = e^x(2x + x^2) = x e^x(2 + x)$ .

Worked example (quotient). $y = (2x + 1)/(x^2 + 3)$ .

$y' = \frac{2(x^2 + 3) - (2x + 1)(2x)}{(x^2 + 3)^2} = \frac{2x^2 + 6 - 4x^2 - 2x}{(x^2 + 3)^2} = \frac{-2x^2 - 2x + 6}{(x^2 + 3)^2}$ .

Tangent at $(a, f(a))$ : $y - f(a) = f'(a)(x - a)$ .

Normal: gradient $= -1/f'(a)$ .

Chain: differentiate outer, KEEP inner, times derivative of inner.
Product/quotient: pick u, v carefully and stay neat.
Tangent and normal: perpendicular gradients multiply to $-1$ .

Quick recap

Limit defines the derivative.
First principles: $\lim_{h \to 0} [f(x+h) - f(x)]/h$ .
Memorise the derivative bank.
Chain, product, quotient rules — and tangent/normal lines.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

$f'(x) = \lim_{h \to 0} [f(x+h) - f(x)]/h$
$d/dx(\sin x) = \cos x$ , $d/dx(\cos x) = -\sin x$
$d/dx(\tan x) = \sec^2 x$
$d/dx(e^x) = e^x$ , $d/dx(\ln x) = 1/x$
$d/dx(\arctan x) = 1/(1 + x^2)$
Chain: $(f \circ g)' = f'(g) g'$

How it’s examined

Paper 1: rule combinations, first principles for $x^n$ or trig. Paper 2: applied differentiation. Paper 3: deeper derivations or analytic problems.

Sources: IB Diploma Programme Mathematics: Analysis and Approaches subject guide (IBO, official). Last reviewed 2026-05-31.

Take this whole topic with you

Step-by-step worked examples — Differentiation 1

Step-by-step solutions to past-paper-style questions on differentiation 1, written exactly the way a tutor would explain them at the board.

1First principles for $x^3$
Building confidence• limits
Question
Show from first principles that $d/dx(x^3) = 3x^2$ .
Step-by-step solution
1. Step 1
  $\frac{(x+h)^3 - x^3}{h} = \frac{3x^2 h + 3xh^2 + h^3}{h} = 3x^2 + 3xh + h^2$ .
2. Step 2
  $\lim_{h \to 0}(3x^2 + 3xh + h^2) = 3x^2$ .
Answer
$3x^2$ — proved.
2Chain — composite of three
Stretch• chain
Question
Differentiate $y = \ln(\cos(2x))$ .
Step-by-step solution
1. Step 1
  Outer: $\ln u \Rightarrow 1/u$ . Middle: $\cos v \Rightarrow -\sin v$ . Inner: $2x \Rightarrow 2$ .
2. Step 2
  $y' = \frac{1}{\cos(2x)} \cdot (-\sin(2x)) \cdot 2 = -2\tan(2x)$ .
Answer
$y' = -2\tan(2x)$ .
3Combining product and quotient
Stretch• product, quotient
Question
Differentiate $y = (x^2 \sin x)/(e^x)$ .
Step-by-step solution
1. Step 1
  Numerator $u = x^2 \sin x$ . By product rule: $u' = 2x \sin x + x^2 \cos x$ .
2. Step 2
  Denominator $v = e^x$ , $v' = e^x$ .
3. Step 3
  $y' = (u'v - uv')/v^2 = [(2x\sin x + x^2\cos x)e^x - x^2\sin x \cdot e^x]/e^{2x}$ .
4. Step 4
  $= [2x\sin x + x^2\cos x - x^2\sin x]/e^x$ .
Answer
$y' = (2x\sin x + x^2\cos x - x^2\sin x)/e^x$ .
4Tangent and normal at a point
Building confidence• tangent, normal
Question
Find the tangent and normal to $y = x^2 - 2x + 3$ at $x = 2$ .
Step-by-step solution
1. Step 1
  $y(2) = 4 - 4 + 3 = 3$ . Point: $(2, 3)$ .
2. Step 2
  $y' = 2x - 2$ , so $y'(2) = 2$ . Tangent gradient $= 2$ .
3. Step 3
  Tangent: $y - 3 = 2(x - 2) \Rightarrow y = 2x - 1$ .
4. Step 4
  Normal gradient $= -1/2$ . Normal: $y - 3 = -\tfrac{1}{2}(x - 2) \Rightarrow y = -x/2 + 4$ .
Answer
Tangent $y = 2x - 1$ . Normal $y = -x/2 + 4$ .

Key Definitions and Keywords — Differentiation 1

Definitions to memorise and the exact keywords mark schemes credit for differentiation 1 answers — sharpened from recent examiner reports for the 2026 IB DP Maths AA HL sitting.

Derivative
Examiner keyword
$f'(x) = \lim_{h \to 0} [f(x+h) - f(x)]/h$ — instantaneous rate of change.
Chain rule
Examiner keyword
$(f \circ g)'(x) = f'(g(x)) g'(x)$ .
Normal line
Examiner keyword
Line perpendicular to the tangent at the point of contact; gradient $-1/f'(a)$ .

Common Mistakes and Misconceptions — Differentiation 1

The traps other students keep falling into on differentiation 1 questions — taken from recent IB DP Maths AA HL examiner reports and mark schemes — and how to avoid them.

✕ $d/dx(\sin 3x) = \cos 3x$ .
Why it happens
Forgetting the chain rule.
How to avoid it
Inner derivative is 3: answer is $3\cos 3x$ .
✕Writing $(u/v)' = (uv' - u'v)/v^2$ .
Why it happens
Sign swap.
How to avoid it
Mnemonic: 'low d high minus high d low' — $(u'v - uv')/v^2$ .
✕Using the tangent gradient as the normal gradient.
Why it happens
Rushing.
How to avoid it
Normal gradient $= -1/$ tangent gradient (perpendicular).

Differentiation 1 — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Differentiation 1

Short Notes - Differentiation 1

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1First principles for x3x^3x3

2Chain — composite of three

3Combining product and quotient

4Tangent and normal at a point

Derivative

Chain rule

Normal line

✕d/dx(sin⁡3x)=cos⁡3xd/dx(\sin 3x) = \cos 3xd/dx(sin3x)=cos3x.

✕Writing (u/v)′=(uv′−u′v)/v2(u/v)' = (uv' - u'v)/v^2(u/v)′=(uv′−u′v)/v2.

✕Using the tangent gradient as the normal gradient.

Differentiation 1 — frequently asked questions

1First principles for $x^3$

✕ $d/dx(\sin 3x) = \cos 3x$ .

✕Writing $(u/v)' = (uv' - u'v)/v^2$ .