Differentiation

CIE A-Level Mathematics· 9709 P1 Topic 7: Differentiation· 15 min read

1. Definition and First Principles Differentiation★★☆☆☆⏱ 5 min

📘 Definition

Derivative from First Principles

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

The derivative is the limit of the gradient of a secant line between two points on a curve, as the distance between the points approaches zero. This gives the gradient of the tangent at the point.

Example:

For f(x) = x^2, the derivative from first principles is 2x

📐 Worked Example

Find the derivative of $f (x) = 3 x^{2}$ from first principles.

1
First calculate $f (x + h)$ :
2
$f(x+h) = 3(x+h)^2 = 3(x^2 + 2xh + h^2) = 3x^2 + 6xh + 3h^2$
3
Subtract $f (x)$ and factor the result:
4
$f(x+h) - f(x) = (3x^2 + 6xh + 3h^2) - 3x^2 = 6xh + 3h^2 = h(6x + 3h)$
5
Divide by $h$ and take the limit as $h \to 0$ :
6
$\frac{f(x+h) - f(x)}{h} = 6x + 3h, \quad \lim_{h \to 0} (6x + 3h) = 6x$
7
Final result: $f^{'} (x) = 6 x$

Exam tip:

Always show all four steps of the first principles method when requested, marks are awarded for each step.

2. The Power Rule for Differentiation★★★☆☆⏱ 6 min

📘 Definition

Power Rule

\text{If } y = ax^n, \quad \frac{dy}{dx} = anx^{n-1}

A general rule for differentiating any power function, valid for all real constants $a$ and $n$ . This rule can be applied term-by-term to sum of functions.

Example:

$5 x^{3}$ differentiates to $15 x^{2}$

📐 Worked Example

Differentiate $y = 4 x^{3} - \frac{2}{x ^{2}} + 5 x$ with respect to $x$ .

1
Rewrite all terms in the form $a x^{n}$ :
2
$y = 4x^3 - 2x^{-2} + 5x^{\frac{1}{2}}$
3
Apply the power rule term-by-term:
4
$\frac{d}{dx}(4x^3) = 4 \times 3 x^{2} = 12x^2$
5
$\frac{d}{dx}(-2x^{-2}) = (-2) \times (-2) x^{-3} = 4x^{-3}$
6
$\frac{d}{dx}(5x^{\frac{1}{2}}) = 5 \times \frac{1}{2} x^{-\frac{1}{2}} = \frac{5}{2}x^{-\frac{1}{2}}$
7
Rewrite with positive roots/denominators for the final answer:
8
$\frac{dy}{dx} = 12x^2 + \frac{4}{x^3} + \frac{5}{2\sqrt{x}}$

3. Equations of Tangents and Normals★★★☆☆⏱ 6 min

The derivative of a curve at a point $(a, f (a))$ gives the gradient of the tangent to the curve at that point. The normal is perpendicular to the tangent at the point of contact, so its gradient is the negative reciprocal of the tangent gradient: $m_{n} = - \frac{1}{m _{t}}$ .

📐 Worked Example

Find the equation of the normal to the curve $y = x^{3} - 2 x + 1$ at the point $(1, 0)$ .

1
First differentiate the curve to get the gradient function:
2
$\frac{dy}{dx} = 3x^2 - 2$
3
Calculate the tangent gradient at $x = 1$ :
4
$m_t = 3(1)^2 - 2 = 1$
5
Find the normal gradient as the negative reciprocal:
6
$m_n = -\frac{1}{1} = -1$
7
Use the point-gradient form of a straight line to find the equation:
8
$y - y_1 = m(x - x_1) \implies y - 0 = -1(x - 1)$
9
Simplify to get the final answer:
10
$y = -x + 1$

Exam tip:

Double-check if the question asks for a tangent or a normal, this is one of the most common avoidable mistakes in exams.

4. Stationary Points★★★★☆⏱ 7 min

📘 Definition

Stationary Point

A point on a curve where the gradient (derivative) is zero, meaning the tangent is horizontal. There are two types: maximum points (gradient changes from positive to negative) and minimum points (gradient changes from negative to positive).

Example:

The curve $y = x^{2}$ has a minimum stationary point at $(0, 0)$ .

📐 Worked Example

Find and classify the stationary points of $y = x^{3} - 3 x + 2$ .

1
Differentiate to get the first derivative:
2
$\frac{dy}{dx} = 3x^2 - 3$
3
Set derivative equal to zero to find x-coordinates of stationary points:
4
$3x^2 - 3 = 0 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1$
5
Find corresponding y-coordinates:
6
$x=1: y=1 - 3 + 2 = 0 \quad (1, 0)$
7
$x=-1: y=-1 + 3 + 2 = 4 \quad (-1, 4)$
8
Use the second derivative test to classify:
9
$\frac{d^2y}{dx^2} = 6x$
10
For $(1, 0)$ : $\frac{d ^{2} y}{d x ^{2}} = 6 (1) = 6 > 0$ , so it is a minimum. For $(- 1, 4)$ : $\frac{d ^{2} y}{d x ^{2}} = 6 (- 1) = - 6 < 0$ , so it is a maximum.

5. Common Pitfalls

Wrong move:

Differentiating a constant term and leaving the constant in the derivative

Why:

Forgetting that a constant $c = c x^{0}$ , so the power rule gives derivative $0 \times c x^{- 1} = 0$

Correct move:

All constant terms always differentiate to zero, remove them from the final derivative

Wrong move:

Skipping rewriting negative/fractional powers, e.g. writing $\frac{d}{d x} (1/ x) = 1$

Why:

The power rule only works for terms in the form $a x^{n}$ , so reciprocals/roots must be rewritten first

Correct move:

Rewrite $1/ x = x^{- 1}$ , then apply the rule to get derivative $- x^{- 2} = - 1/ x^{2}$

Wrong move:

Using the tangent gradient for the normal line equation

Why:

Confusing the definition of tangent and normal, or forgetting the perpendicular gradient rule

Correct move:

Always calculate the normal gradient as the negative reciprocal of the tangent gradient: $m_{n} = - 1/ m_{t}$

Wrong move:

Reversing the second derivative test: calling a positive second derivative a maximum

Why:

Misremembering the classification rule for stationary points

Correct move:

Positive second derivative = minimum, negative second derivative = maximum

6. Quick Reference Cheatsheet

Concept	Rule/Formula
First principles derivative	f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
Power rule	y = ax^n \implies \frac{dy}{dx} = anx^{n-1}
Tangent gradient at $x = a$	m_t = \frac{dy}{dx} \bigg\|_{x=a}
Normal gradient at $x = a$	m_n = -\frac{1}{m_t}
Stationary point condition	\frac{dy}{dx} = 0
Second derivative test	$d^{2} y / d x^{2} > 0$ : minimum; $< 0$ : maximum

7. Frequently Asked

Do I always need to use first principles for differentiation?

No, only when the question explicitly asks for a proof from first principles. All other questions allow you to use the power rule directly.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
First principles differentiation
2023 · 1
Tangent and normal question
2024 · 1
Stationary points classification

What's Next

Differentiation is the foundation of all further calculus in A-Level Mathematics. After mastering this P1 introduction, you will extend these concepts to trigonometric, exponential and logarithmic functions in Pure Mathematics 2, learn new rules for differentiating products and quotients of functions, and apply differentiation to connected rates of change and optimisation problems. These core topics build directly on the basic rules and skills you learned in this sub-topic, so mastering this content is critical for success in the full A-Level.