# Differentiation (Pure Mathematics 4)

> Edexcel International A-Level Mathematics · IAL P4 Maths
> Source: https://www.owlsprep.com/study/edexcel-ial-math-p4-differentiation/

This guide covers all Edexcel IAL P4 differentiation content: implicit and parametric differentiation, tangent/normal equations for these curve types, and formation of differential equations including connected rates problems.

**Prerequisites:** [Differentiation rules (chain, product, quotient) from P3](https://www.owlsprep.com/study/edexcel-ial-math-p3-differentiation/); [Tangent and normal equations for Cartesian curves from P1](https://www.owlsprep.com/study/edexcel-ial-math-p1-coordinate-geometry/)

## Learning objectives

- Differentiate implicitly defined functions to find gradients, tangents and normals
- Differentiate parametrically defined functions to find gradients, tangents and normals
- Form simple differential equations, including for connected rates of change problems

## Implicit Differentiation & Tangent/Normal Equations

**Implicit Function** — A function where x and y are not separated (y is not isolated on one side of the equation), e.g. $x^2 + 3xy + y^2 = 10$

*Example:* $x^2 + 2y^2 - xy = 2$

To differentiate implicitly, apply standard P3 differentiation rules to all terms with respect to $x$, and use the chain rule on terms containing $y$: multiply by $\frac{dy}{dx}$ every time you differentiate a term in $y$. Rearrange the resulting equation to isolate $\frac{dy}{dx}$, then substitute coordinate values to find the gradient at a given point.

**Worked example:** Find the equation of the tangent to the curve $x^2 + 2y^2 - xy = 2$ at the point (1, 1).

1. Differentiate each term with respect to $x$:

   $$\frac{d}{dx}(x^2) + \frac{d}{dx}(2y^2) - \frac{d}{dx}(xy) = \frac{d}{dx}(4)$$
2. Apply chain and product rules to expand:

   $$2x + 4y\frac{dy}{dx} - (y + x\frac{dy}{dx}) = 0$$
3. Rearrange to isolate $\frac{dy}{dx}$:

   $$\frac{dy}{dx} = \frac{y - 2x}{4y - x}$$
4. Substitute $x=1, y=1$ to find the tangent gradient:

   $$m = \frac{1 - 2}{4 - 1} = -\frac{1}{3}$$
5. Use point-gradient form to write the tangent equation:

   $$y - 1 = -\frac{1}{3}(x - 1) \implies x + 3y = 4$$

> **Exam tip:** Always isolate $\frac{dy}{dx}$ before substituting coordinate values to avoid arithmetic errors and save time.

*Calculator:* allowed

## Parametric Differentiation & Tangent/Normal Equations

**Parametric Function** — A curve where $x$ and $y$ are both defined in terms of a third independent variable (parameter, often $t$), e.g. $x = 2t + 1$, $y = t^2 - 3t$

For parametric equations $x = f(t)$, $y = g(t)$, the gradient of the curve is given by $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, provided $\frac{dx}{dt} \neq 0$. If $\frac{dx}{dt}=0$ the tangent is vertical; if $\frac{dy}{dt}=0$ the tangent is horizontal.

**Worked example:** A curve is defined parametrically by $x = t^2 + 1$, $y = 3t - t^3$. Find the equation of the normal to the curve at the point where $t=1$.

1. Differentiate $x$ and $y$ with respect to $t$:

   $$\frac{dx}{dt} = 2t, \frac{dy}{dt} = 3 - 3t^2$$
2. Calculate $\frac{dy}{dx}$ using the parametric gradient rule:

   $$\frac{dy}{dx} = \frac{3 - 3t^2}{2t}$$
3. Substitute $t=1$: $\frac{dy}{dx} = 0$, so the tangent is horizontal, meaning the normal is vertical.
4. Find coordinates at $t=1$: $x=1^2 +1=2$, $y=3(1) - 1^3 = 2$
5. Write the equation of the vertical normal through (2, 2):

   $$x = 2$$

> **Exam tip:** For vertical/horizontal tangents, always state the gradient type explicitly rather than just writing 0 or undefined, to demonstrate you understand the context.

*Calculator:* allowed

## Formation of Simple Differential Equations

**Differential Equation** — An equation that relates a function to one or more of its derivatives, e.g. $\frac{dP}{dt} = kP$

You will be given a contextual or algebraic relationship between variables, and asked to form a differential equation by translating verbal statements to derivative terms. You do **not** need to solve these equations in this topic. Always define any proportionality constants you introduce.

**Worked example:** The rate of decrease of the mass $m$ of a radioactive substance is proportional to the current mass. Form a differential equation to model this relationship.

1. Translate "rate of decrease of mass" to $-\frac{dm}{dt}$, where $t$ is time (the negative sign indicates decrease).
2. Translate "proportional to current mass" to $km$, where $k$ is a positive constant of proportionality.
3. Combine to form the final differential equation:

   $$-\frac{dm}{dt} = km \implies \frac{dm}{dt} = -km$$

> **Exam tip:** You will lose 1 mark if you do not define any constants you introduce in your differential equation.

*Calculator:* allowed

## Connected Rates of Change Problems

Connected rates problems involve linking the rate of change of one quantity to the rate of change of another, using the chain rule: $\frac{da}{db} = \frac{da}{dc} \times \frac{dc}{db}$. You will often need to use standard geometry formulas (e.g. volume of a sphere, area of a circle) to link the two quantities.

**Worked example:** The radius $r$ of a spherical balloon increases at a constant rate of 2 cm/s. Find the rate of increase of the volume $V$ of the balloon when $r = 5$ cm. Use the formula $V = \frac{4}{3}\pi r^3$.

1. Identify given rate: $\frac{dr}{dt} = 2$ cm/s. We need to find $\frac{dV}{dt}$ when $r=5$.
2. Apply the chain rule to link the two rates:

   $$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$$
3. Differentiate the volume formula with respect to $r$:

   $$\frac{dV}{dr} = 4\pi r^2$$
4. Substitute $r=5$ and $\frac{dr}{dt}=2$ to calculate the final rate:

   $$\frac{dV}{dt} = 4\pi (5)^2 \times 2 = 200\pi \text{ cm}^3/\text{s}$$

> **Exam tip:** Always include units in your final answer for rates questions, as 1 mark is usually allocated for correct units.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to multiply differentiated $y$ terms by $\frac{dy}{dx}$ when doing implicit differentiation, e.g. writing $\frac{d}{dx}(y^2) = 2y$.
  - Why it fails: Implicit differentiation requires the chain rule, as $y$ is a function of $x$, so you must account for the derivative of $y$ with respect to $x$.
  - Correct: Multiply every differentiated $y$ term by $\frac{dy}{dx}$ immediately when applying the rule.
- **Wrong:** Swapping the numerator and denominator for parametric differentiation, writing $\frac{dy}{dx} = \frac{dx/dt}{dy/dt}$.
  - Why it fails: The chain rule gives $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ exclusively, and swapping gives the reciprocal of the correct gradient.
  - Correct: Confirm the order of derivatives before performing division, checking that $dt$ cancels out when multiplying the fractions.
- **Wrong:** Adding a constant of integration when forming differential equations.
  - Why it fails: Constants of integration only apply when solving differential equations, not when forming them from a given relationship.
  - Correct: Omit integration constants during formation, only define proportionality constants if required by the problem.
- **Wrong:** Using the wrong chain rule structure for connected rates, e.g. writing $\frac{dV}{dt} = \frac{dr}{dt} \div \frac{dV}{dr}$.
  - Why it fails: Incorrect chain rule application leads to inverted units and numerically wrong answers.
  - Correct: Write the required rate first, then express it as a product of two known/derivable rates that cancel the intermediate variable.
- **Wrong:** Calculating the normal gradient as the same as the tangent gradient, or using a positive reciprocal instead of negative.
  - Why it fails: The normal is perpendicular to the tangent, so their gradients multiply to -1 for non-zero finite values.
  - Correct: For tangent gradient $m$, normal gradient is $-\frac{1}{m}$; handle horizontal/vertical cases separately where this formula does not apply.

## Cheatsheet

| Concept | Formula / Method | Key Exam Note |
| --- | --- | --- |
| Implicit Differentiation | Differentiate all terms w.r.t $x$, multiply $y$ terms by $\frac{dy}{dx}$, rearrange for $\frac{dy}{dx}$ | Apply product/quotient rules to mixed $x,y$ terms |
| Parametric Differentiation | $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | Gradient undefined if $\frac{dx}{dt}=0$ (vertical tangent) |
| Differential Equation Formation | Translate verbal rate statements to derivative terms, add proportionality constants | Do NOT solve the equation for this topic |
| Connected Rates | Use chain rule to link required rate to given rate, e.g. $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$ | Always include units in your final answer |

## What's next

Now that you have mastered P4 differentiation skills, you are ready to apply these to the P4 Integration topic, where you will use differentiation results to reverse-engineer integrals, and solve the separable differential equations you have learned to form here. These differentiation skills also frequently appear in longer 6-8 mark P4 exam questions combined with coordinate geometry and modelling tasks, so practicing past paper questions is critical to build speed and accuracy. If you are studying Further Mathematics, these skills are also foundational for further calculus topics including polar coordinates and differential equations.

- [P3 Differentiation Rules (Revision)](https://www.owlsprep.com/study/edexcel-ial-math-p3-differentiation/)

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