Differentiation

Q: Do I need to differentiate from first principles in 9709 P1?

No. Formal differentiation from first principles is not required in 9709 Paper 1. You should understand the derivative as the limit of the gradients of a sequence of chords (which gives the gradient of the tangent), but in practice you differentiate using the power rule and the chain rule.

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

1. Definition and First Principles Differentiation★☆☆☆☆⏱ 3 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

As the second point of a chord slides toward the first, the sequence of chord gradients approaches a limiting value — the gradient of the tangent at that point. This limit is the derivative $f^{'} (x)$ . In 9709 Paper 1 you need to understand this idea, but you are not required to differentiate from first principles; in practice you use the power rule and chain rule that follow.

Exam tip:

You are not required to differentiate from first principles in 9709 P1. Understand the derivative as the limiting gradient of a sequence of chords, then differentiate using the power and chain rules.

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 any real constant $a$ and any rational power $n$ . This rule can be applied term-by-term to a 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. The Chain Rule★★★☆☆⏱ 6 min

📘 Definition

Chain Rule

\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

To differentiate a composite function $y = f (g (x))$ , substitute $u = g (x)$ so that $y$ becomes a function of $u$ . Differentiate the outer function with respect to $u$ , then multiply by the derivative of the inner function — differentiate the outside, then multiply by the derivative of the inside.

Example:

For $y = (2 x + 1)^{5}$ , let $u = 2 x + 1$ : $\frac{d y}{d x} = 5 u^{4} \times 2 = 10 (2 x + 1)^{4}$ .

📐 Worked Example

Differentiate $y = 3 x^{2} - 1$ with respect to $x$ .

1
Write the function as a power and identify the inner function:
2
$y = (3x^2 - 1)^{\frac{1}{2}}, \quad u = 3x^2 - 1$
3
Differentiate the outside with respect to $u$ :
4
$\frac{dy}{du} = \frac{1}{2} u^{-\frac{1}{2}} = \frac{1}{2}(3x^2 - 1)^{-\frac{1}{2}}$
5
Differentiate the inside with respect to $x$ :
6
$\frac{du}{dx} = 6x$
7
Multiply the two derivatives (chain rule):
8
$\frac{dy}{dx} = \frac{1}{2}(3x^2 - 1)^{-\frac{1}{2}} \times 6x = \frac{3x}{\sqrt{3x^2 - 1}}$

Exam tip:

For a bracket raised to a power, $\frac{d}{d x} (a x + b)^{n} = n (a x + b)^{n - 1} \times a$ . Always include the derivative of the inside.

4. 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.

5. Increasing and Decreasing Functions★★★☆☆⏱ 5 min

📘 Definition

Increasing and Decreasing Functions

f'(x) > 0 \implies \nearrow, \quad f'(x) < 0 \implies \searrow

A function is increasing on an interval where its derivative is positive, and decreasing on an interval where its derivative is negative. The sign of $f^{'} (x)$ tells you whether the curve is going up or down as $x$ increases.

Example:

For $f (x) = x^{2}$ , $f^{'} (x) = 2 x$ : the function is decreasing for $x < 0$ and increasing for $x > 0$ .

📐 Worked Example

Find the values of $x$ for which $f (x) = x^{3} - 3 x^{2}$ is decreasing.

1
Differentiate to find the gradient function:
2
$f'(x) = 3x^2 - 6x$
3
A function is decreasing where $f^{'} (x) < 0$ :
4
$3x^2 - 6x < 0 \implies 3x(x - 2) < 0$
5
Solve the inequality (the critical values are $x = 0$ and $x = 2$ ):
6
$0 < x < 2$
7
So $f (x)$ is decreasing for $0 < x < 2$ .

Exam tip:

State your answer as an inequality or interval. A function is increasing where $f^{'} (x) > 0$ and decreasing where $f^{'} (x) < 0$ — do not get these the wrong way round.

6. 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.

7. Rates of Change and Connected Rates★★★★☆⏱ 6 min

A derivative is a rate of change: $\frac{d y}{d x}$ measures how fast $y$ changes with $x$ . When two quantities both change with time, their rates are connected by the chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$ . This lets you find an unknown rate from a known one.

📐 Worked Example

The radius $r$ cm of a circle increases at a constant rate of $0.2$ cm s $^{- 1}$ . Find the rate at which the area $A$ is increasing at the instant when $r = 5$ cm.

1
Write down the known rate and the relationship between $A$ and $r$ :
2
$\frac{dr}{dt} = 0.2, \quad A = \pi r^2$
3
Differentiate $A$ with respect to $r$ :
4
$\frac{dA}{dr} = 2\pi r$
5
Connect the rates using the chain rule:
6
$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 2\pi r \times 0.2$
7
Substitute $r = 5$ :
8
$\frac{dA}{dt} = 2\pi (5)(0.2) = 2\pi \approx 6.28 \text{ cm}^2\,\text{s}^{-1}$

Exam tip:

Identify the rate you are given and the rate you want, then build the chain of derivatives that links them. Include units in your final answer.

8. 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

9. 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}
Chain rule	\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}
Tangent gradient at $x = a$	m_t = \frac{dy}{dx} \bigg\|_{x=a}
Normal gradient at $x = a$	m_n = -\frac{1}{m_t}
Increasing / decreasing	$f^{'} (x) > 0$ : increasing; $f^{'} (x) < 0$ : decreasing
Stationary point condition	\frac{dy}{dx} = 0
Second derivative test	$d^{2} y / d x^{2} > 0$ : minimum; $< 0$ : maximum
Connected rates of change	\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}

10. Frequently Asked

Do I need to differentiate from first principles in 9709 P1?

No. Formal differentiation from first principles is not required in 9709 Paper 1. You should understand the derivative as the limit of the gradients of a sequence of chords (which gives the gradient of the tangent), but in practice you differentiate using the power rule and the chain rule.

What's Next

Differentiation is the foundation of all further calculus in A-Level Mathematics. Within this P1 topic you have already met the chain rule, increasing and decreasing functions, connected rates of change, and optimisation using stationary points. In Pure Mathematics 2 and 3 you will extend differentiation to trigonometric, exponential and logarithmic functions, and learn new rules for differentiating products and quotients of functions. Mastering the P1 rules and skills here is essential before taking on those later techniques.