# Differentiation

> CIE A-Level Mathematics · Pure Mathematics 1 (P1)
> Source: https://www.owlsprep.com/study/cie-9709-u1-differentiation/

This sub-topic introduces core differentiation concepts for CIE A-Level P1. You will learn key rules for differentiating power functions, find tangents and normals, and locate and classify stationary points, a frequent high-weight exam topic.

**Prerequisites:** [Coordinate geometry and straight lines](https://www.owlsprep.com/study/cie-9709-u1-coordinate-geometry/); [Polynomials and indices rules](https://www.owlsprep.com/study/cie-9709-u1-polynomials/)

## Learning objectives

- Understand differentiation as the gradient of a tangent to a curve
- Apply the power rule to differentiate polynomial, negative and fractional power functions
- Find equations of tangents and normals to polynomial curves
- Locate and classify stationary points of polynomial functions
- Differentiate composite functions using the chain rule
- Determine intervals where a function is increasing or decreasing using $f'(x)$
- Apply differentiation to rates of change and connected rates of change
- Find and use the second derivative $\frac{d^2y}{dx^2}$ to classify stationary points

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P1 past papers (2016–2025): **Differentiation** appears **162 times** in the last 10 years — **21.2%** of all P1 questions (162 of 692).

Most-tested forms: Stationary points (max/min) (68), Tangent and normal lines (54), Second derivative test (29), Rate of change / related rates (28), Increasing and decreasing functions (20).

## Definition and First Principles Differentiation

**Derivative from First Principles** — 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.

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

*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.

## The Power Rule for Differentiation

**Power Rule** — 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.

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

*Example:* $5x^3$ differentiates to $15x^2$

> **tip**
>
> Always rewrite reciprocals and roots as negative and fractional powers before applying the power rule. This avoids common index errors.

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

1. Rewrite all terms in the form $ax^n$:
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. $$\frac{dy}{dx} = 12x^2 + \frac{4}{x^3} + \frac{5}{2\sqrt{x}}$$

## The Chain Rule

**Chain Rule** — 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.

*Notation:* \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}

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

> **tip**
>
> The most common slip is forgetting to multiply by the derivative of the inside. For $(2x+1)^5$ the inner derivative is $2$, so the answer is $10(2x+1)^4$, not $5(2x+1)^4$.

**Worked example:** Differentiate $y = \sqrt{3x^2 - 1}$ 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, $\frac{d}{dx}(ax+b)^n = n(ax+b)^{n-1} \times a$. Always include the derivative of the inside.

## Equations of Tangents and Normals

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 $y = x^3 - 2x + 1$ 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.

## Increasing and Decreasing Functions

**Increasing and Decreasing Functions** — 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.

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

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

**Worked example:** Find the values of $x$ for which $f(x) = x^3 - 3x^2$ 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.

## Stationary Points

**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 $y = x^3 - 3x + 2$.

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 $(1, 0)$: $\frac{d^2y}{dx^2} = 6(1) = 6 > 0$, so it is a minimum. For $(-1, 4)$: $\frac{d^2y}{dx^2} = 6(-1) = -6 < 0$, so it is a maximum.

## Rates of Change and Connected Rates

A derivative is a rate of change: $\frac{dy}{dx}$ measures how fast $y$ changes with $x$. When two quantities both change with time, their rates are connected by the chain rule: $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$. 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.

## Common pitfalls

- **Wrong:** Differentiating a constant term and leaving the constant in the derivative
  - Why it fails: Forgetting that a constant $c = c x^0$, so the power rule gives derivative $0 \times c x^{-1} = 0$
  - Correct: All constant terms always differentiate to zero, remove them from the final derivative
- **Wrong:** Skipping rewriting negative/fractional powers, e.g. writing $\frac{d}{dx}(1/x) = 1$
  - Why it fails: The power rule only works for terms in the form $ax^n$, so reciprocals/roots must be rewritten first
  - Correct: Rewrite $1/x = x^{-1}$, then apply the rule to get derivative $-x^{-2} = -1/x^2$
- **Wrong:** Using the tangent gradient for the normal line equation
  - Why it fails: Confusing the definition of tangent and normal, or forgetting the perpendicular gradient rule
  - Correct: Always calculate the normal gradient as the negative reciprocal of the tangent gradient: $m_n = -1/m_t$
- **Wrong:** Reversing the second derivative test: calling a positive second derivative a maximum
  - Why it fails: Misremembering the classification rule for stationary points
  - Correct: Positive second derivative = minimum, negative second derivative = maximum

## 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 | $d^2y/dx^2 > 0$: minimum; $<0$: maximum |
| Connected rates of change | \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} |

## 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.

- [Integration (Reverse of Differentiation)](https://www.owlsprep.com/study/cie-9709-u1-integration/)
- [Pure Mathematics 3](https://www.owlsprep.com/study/cie-9709-u2-overview/)

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