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 and analyzing the behavior of functions.

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 fundamental in IB DP Mathematics for simplifying the process of differentiation.

Why is understanding the derivative important in IB DP Mathematics?

Understanding the derivative is crucial in IB DP Mathematics because it allows students to analyze and interpret the behavior of functions. It helps in solving problems related to rates of change, optimization, and motion, which are essential in both theoretical and applied mathematics contexts.

What are common mistakes to avoid when learning Differentiation 1?

Common mistakes in Differentiation 1 include misapplying the power rule, forgetting to apply the chain rule when necessary, and neglecting to simplify the derivative expression. It's important to carefully follow differentiation rules and practice regularly to avoid these errors in IB DP Mathematics.

How does Differentiation 1 relate to real-world applications?

Differentiation 1 is directly applicable to real-world scenarios such as calculating speed, acceleration, and optimizing business processes. In IB DP Mathematics, understanding these applications helps students connect mathematical concepts to practical situations, enhancing their problem-solving skills.

Why is "Trying to differentiate $\dfrac{1}{x}$ by 'eyeballing' instead of rewriting as $x^{-1}$." a common mistake in Differentiation 1?

Why it happens: Rushing. How to avoid it: ALWAYS rewrite as a power before applying the power rule. 1/x^n = x^-n, √[n]x = x^1/n.

Why is "Writing the tangent as $y = f'(a) x + c$ without computing $f(a)$." a common mistake in Differentiation 1?

Why it happens: Forgetting that the tangent passes through the point on the curve. How to avoid it: Use the point-gradient form: y - f(a) = f'(a)(x - a). Both f(a) and f'(a) are needed.

Why is "Using $f'(a)$ as the normal's gradient." a common mistake in Differentiation 1?

Why it happens: Mixing up the two lines. How to avoid it: Normal gradient = -1/f'(a) — negative reciprocal. (When f'(a) = 0 the normal is vertical: x = a.)

DifferentiationIB Maths AA SL

Differentiation 1

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Differentiation 1 — IB Maths AA SL: limits, gradient as a function, power rule, and tangents

AA SL calculus begins here: the derivative as the gradient of the tangent, the power rule, and using $f'(x)$ to find tangent and normal lines. This note covers the foundations for ALL AA SL calculus work.

At a glance

DERIVATIVE: $f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ — gradient of the tangent at $x$ .
POWER RULE: $\dfrac{d}{dx}(x^n) = n x^{n-1}$ for any $n \in \mathbb{R}$ .
CONSTANT and SUM rules: $\dfrac{d}{dx}(c) = 0$ ; $\dfrac{d}{dx}(f + g) = f' + g'$ .
EXPONENTIAL: $\dfrac{d}{dx}(e^x) = e^x$ .
LOGARITHM: $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ .
TRIG: $\dfrac{d}{dx}(\sin x) = \cos x$ ; $\dfrac{d}{dx}(\cos x) = -\sin x$ .
TANGENT at $(a, f(a))$ : $y - f(a) = f'(a)(x - a)$ .

What you’ll learn

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

AO1 — State the power rule and the derivatives of $e^x$ , $\ln x$ , $\sin x$ , $\cos x$ .
AO1 — Differentiate sums of standard functions.
AO2 — Find the gradient at a point and equations of tangent and normal lines.
AO2 — Use first principles to derive simple derivatives.
AO3 — Use Leibniz $\left(\dfrac{dy}{dx}\right)$ and prime $(f'(x))$ notation correctly.

Derivative as a limit

The slope of the secant becomes the slope of the tangent.

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

This is the slope of the SECANT line from $(x, f(x))$ to $(x + h, f(x + h))$ , as $h$ shrinks to 0.

Worked example (first principles). Derive $\dfrac{d}{dx}(x^2)$ .

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

(So the power rule checks out for $n = 2$ .)

Notation. Two equivalent forms — both appear in mark schemes:

Lagrange: $f'(x)$ or $y' =$ .
Leibniz: $\dfrac{dy}{dx}$ or $\dfrac{d}{dx}\Big(f(x)\Big)$ .

Geometric meaning. $f'(a)$ = slope of the tangent line to $y = f(x)$ at $x = a$ .

The tangent line touches the curve at the point and has the same slope as the curve there. The derivative captures that local slope.

$f'(x)$ is a NEW function, giving the slope at each $x$ .
Both Lagrange and Leibniz notations are accepted.
First principles is examined in Paper 1 short proofs.

Power, exponential, logarithm and trig rules

Memorise these — they're in the formula booklet.

Power rule: $\dfrac{d}{dx}(x^n) = n x^{n-1}$ , valid for any real $n$ — including negative and fractional.

Worked examples.

$f(x)$	$f'(x)$
$x^7$	$7x^6$
$x^{-3} = \dfrac{1}{x^3}$	$-3x^{-4} = -\dfrac{3}{x^4}$
$\sqrt{x} = x^{1/2}$	$\dfrac{1}{2} x^{-1/2} = \dfrac{1}{2\sqrt{x}}$
$\dfrac{4}{x^2} = 4x^{-2}$	$-8 x^{-3} = -\dfrac{8}{x^3}$

Sum/difference and constants.

$\dfrac{d}{dx}(c) = 0$ for any constant $c$ .
$\dfrac{d}{dx}(c f(x)) = c f'(x)$ .
$\dfrac{d}{dx}(f + g) = f' + g'$ .

Exponential. $\dfrac{d}{dx}(e^x) = e^x$ .

Natural log. $\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ .

Trig. $\dfrac{d}{dx}(\sin x) = \cos x$ , $\dfrac{d}{dx}(\cos x) = -\sin x$ , $\dfrac{d}{dx}(\tan x) = \sec^2 x$ .

All of these are IN THE FORMULA BOOKLET. They're not memorisation challenges; they're notation challenges.

Worked example. Differentiate $f(x) = 3x^4 - \dfrac{2}{x} + 5\sin x - \ln x$ .

Rewrite $\dfrac{2}{x}$ as $2x^{-1}$ to use the power rule.

$f'(x) = 12 x^3 - (-2 x^{-2}) + 5\cos x - \dfrac{1}{x} = 12x^3 + \dfrac{2}{x^2} + 5\cos x - \dfrac{1}{x}.$

Power rule works for ANY real exponent.
Always rewrite $\dfrac{a}{x^n}$ as $a x^{-n}$ before differentiating.
Sum/difference of derivatives — linearity.

Tangent and normal lines

$f'(a)$ is the gradient; normal has gradient $-1/f'(a)$ .

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

Normal line (perpendicular to tangent) at the same point: $y - f(a) = -\frac{1}{f'(a)} (x - a) \quad (\text{provided } f'(a) \ne 0).$

Worked example. Find the equations of the tangent and normal to $y = x^3 - 4x + 5$ at $x = 1$ .

$y(1) = 1 - 4 + 5 = 2$ . Point: $(1, 2)$ .

$y' = 3x^2 - 4$ . $y'(1) = -1$ . Gradient of tangent = $-1$ .

Tangent: $y - 2 = -1(x - 1) \Rightarrow y = -x + 3$ .

Normal: gradient $= 1$ (negative reciprocal of $-1$ ).

Normal: $y - 2 = 1(x - 1) \Rightarrow y = x + 1$ .

Strategy (do every time):

Find $y$ -coordinate of the point.
Differentiate.
Evaluate $f'(a)$ at the $x$ -value.
Write tangent equation in point-gradient form.
Take negative reciprocal for the normal.

This is the bread-and-butter of AA SL calculus — appears in every Paper 1 and Paper 2.

Tangent gradient = $f'(a)$ .
Normal gradient = $-\dfrac{1}{f'(a)}$ .
Always evaluate $f(a)$ AND $f'(a)$ separately.
Point-gradient form is fastest.

How it’s examined

Paper 1: differentiate sums, find tangent / normal at a stated point. Paper 2: GDC for evaluating $f'(a)$ when algebra is messy. Examiner reports stress that students who rewrite $\dfrac{1}{x}$ as $x^{-1}$ before differentiating make far fewer mistakes.

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 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.

1Differentiate using the power rule
Getting started• power rule
Question
Find $\dfrac{dy}{dx}$ for $y = 5x^4 - \dfrac{3}{x^2} + 7\sqrt{x}$ .
Step-by-step solution
1. Step 1
  Rewrite using powers: $y = 5x^4 - 3x^{-2} + 7x^{1/2}$ .
2. Step 2
  Differentiate: $\dfrac{dy}{dx} = 20x^3 + 6x^{-3} + \dfrac{7}{2}x^{-1/2}$ .
3. Step 3
  Rewrite: $20x^3 + \dfrac{6}{x^3} + \dfrac{7}{2\sqrt{x}}$ .
Answer
$\dfrac{dy}{dx} = 20x^3 + \dfrac{6}{x^3} + \dfrac{7}{2\sqrt{x}}$ .
2Combination — trig + exponential
Building confidence• trig, exponential
Question
Differentiate $f(x) = 3e^x - 4\sin x + 2\ln x$ .
Step-by-step solution
1. Step 1
  Use linearity: $f'(x) = 3 \dfrac{d}{dx}(e^x) - 4 \dfrac{d}{dx}(\sin x) + 2 \dfrac{d}{dx}(\ln x)$ .
2. Step 2
  Standard derivatives: $f'(x) = 3e^x - 4\cos x + \dfrac{2}{x}$ .
Answer
$f'(x) = 3e^x - 4\cos x + \dfrac{2}{x}$ .
3Tangent equation
Building confidence• tangent
Question
Find the tangent to $y = x^2 - 5x + 3$ at $x = 2$ .
Step-by-step solution
1. Step 1
  Point: $y(2) = 4 - 10 + 3 = -3$ , so $(2, -3)$ .
2. Step 2
  $y' = 2x - 5$ , $y'(2) = -1$ .
3. Step 3
  Tangent: $y - (-3) = -1(x - 2) \Rightarrow y = -x - 1$ .
Answer
$y = -x - 1$ .
4Tangent and normal
Stretch• normal
Question
Find the tangent and normal to $y = e^x$ at $x = 0$ .
Step-by-step solution
1. Step 1
  Point: $(0, e^0) = (0, 1)$ .
2. Step 2
  $y'(x) = e^x$ , $y'(0) = 1$ .
3. Step 3
  Tangent: $y - 1 = 1(x - 0) \Rightarrow y = x + 1$ .
4. Step 4
  Normal gradient = $-1$ . Normal: $y - 1 = -1(x - 0) \Rightarrow y = -x + 1$ .
Answer
Tangent $y = x + 1$ ; normal $y = -x + 1$ .

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 SL sitting.

Derivative $f'(x)$
Examiner keyword
The function $f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ , giving the slope of the tangent at each $x$ .
Tangent line
Examiner keyword
Line touching the curve at one point with the same gradient as the curve there.
Normal line
Line PERPENDICULAR to the tangent at the point of contact.

Common Mistakes and Misconceptions — Differentiation 1

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

✕Trying to differentiate $\dfrac{1}{x}$ by 'eyeballing' instead of rewriting as $x^{-1}$ .
Why it happens
Rushing.
How to avoid it
ALWAYS rewrite as a power before applying the power rule. $\dfrac{1}{x^n} = x^{-n}$ , $\sqrt[n]{x} = x^{1/n}$ .
✕Writing the tangent as $y = f'(a) x + c$ without computing $f(a)$ .
Why it happens
Forgetting that the tangent passes through the point on the curve.
How to avoid it
Use the point-gradient form: $y - f(a) = f'(a)(x - a)$ . Both $f(a)$ and $f'(a)$ are needed.
✕Using $f'(a)$ as the normal's gradient.
Why it happens
Mixing up the two lines.
How to avoid it
Normal gradient $= -\dfrac{1}{f'(a)}$ — negative reciprocal. (When $f'(a) = 0$ the normal is vertical: $x = a$ .)

Past paper style quiz

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Differentiation 1 — frequently asked questions

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

Differentiation 1

Short Study Notes in the form of Slides

Short Study Notes — Differentiation 1

Detailed Notes

What you’ll learn

How it’s examined

1Differentiate using the power rule

2Combination — trig + exponential

3Tangent equation

4Tangent and normal

Derivative f′(x)f'(x)f′(x)

Tangent line

Normal line

✕Trying to differentiate 1x\dfrac{1}{x}x1​ by 'eyeballing' instead of rewriting as x−1x^{-1}x−1.

✕Writing the tangent as y=f′(a)x+cy = f'(a) x + cy=f′(a)x+c without computing f(a)f(a)f(a).

✕Using f′(a)f'(a)f′(a) as the normal's gradient.

Past paper style quiz

Differentiation 1 — frequently asked questions

Derivative $f'(x)$

✕Trying to differentiate $\dfrac{1}{x}$ by 'eyeballing' instead of rewriting as $x^{-1}$ .

✕Writing the tangent as $y = f'(a) x + c$ without computing $f(a)$ .

✕Using $f'(a)$ as the normal's gradient.