Differentiation (Edexcel IAL Pure Mathematics 1)

Mathematics· P1 4.1–4.3· 20 min read

1. What is the Derivative?★★☆☆☆⏱ 4 min

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📘 Definition

Derivative

$f^{'} (x)$ or $\frac{d y}{d x}$

The gradient of the tangent to the curve $y = f (x)$ at a given point, representing the instantaneous rate of change of $y$ with respect to $x$ .

Example:

If $f (t)$ models distance travelled over time, $f^{'} (t)$ is instantaneous speed.

Differentiation from first principles uses the informal concept of a limit: you take two points on a curve separated by a small increment $h$ , calculate the gradient of the chord between them, then let $h$ approach 0 to get the tangent gradient. For P1, you only need to understand this concept, not prove derivative rules from first principles.

📘 Definition

Second Derivative

$f^{''} (x)$ or $\frac{d ^{2} y}{d x ^{2}}$

The result of differentiating the first derivative of a function, representing the rate of change of the gradient of the curve.

📐 Worked Example

State what $f^{'} (t)$ and $f^{''} (t)$ represent for the function $f (t) = 3 t + 2 t^{2}$ , which models the height of a ball in metres at time $t$ seconds.

1
$f^{'} (t)$ is the first derivative of height with respect to time, so it represents the instantaneous vertical velocity of the ball in $m s^{- 1}$ .
2
$f^{''} (t)$ is the second derivative of height with respect to time, so it represents the instantaneous vertical acceleration of the ball in $m s^{- 2}$ .

Exam tip:

If asked to explain the meaning of a derivative in context, always include units where possible to gain full marks.

2. Differentiating Powers of x★★★☆☆⏱ 6 min

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The core differentiation rule you need for P1 applies to any term of the form $a x^{n}$ , where $a$ is a constant and $n$ is any rational number (positive, negative, fractional).

📘 Definition

Power Rule for Differentiation

For $f (x) = a x^{n}$ , the first derivative is $f^{'} (x) = an x^{n - 1}$ . To differentiate a sum or difference of terms, differentiate each term individually and add/subtract the results.

Before differentiating any function, first simplify or expand it into a sum of separate $a x^{n}$ terms. You do NOT need product, quotient or chain rules for P1: all given functions can be rewritten this way.

📐 Worked Example

Differentiate $y = 4 x^{3} + 2 x - \frac{3}{x ^{2}}$ with respect to $x$ , then find the second derivative $\frac{d ^{2} y}{d x ^{2}}$ .

1
Rewrite all terms in $a x^{n}$ form first:
2
$y = 4x^3 + 2x^{\frac{1}{2}} - 3x^{-2}$
3
Apply the power rule to each term for the first derivative:
4
$\frac{dy}{dx} = (3 \times 4)x^{2} + (\frac{1}{2} \times 2)x^{-\frac{1}{2}} - (-2 \times 3)x^{-3}$
5
Simplify the first derivative:
6
$\frac{dy}{dx} = 12x^2 + x^{-\frac{1}{2}} + 6x^{-3}$
7
Differentiate again to get the second derivative, applying the power rule to each term of the first derivative:
8
$\frac{d^2y}{dx^2} = 24x - \frac{1}{2}x^{-\frac{3}{2}} - 18x^{-4}$

Exam tip:

Always rewrite square roots, cube roots and terms with $x$ in the denominator into index form before differentiating to avoid arithmetic mistakes.

3. Equations of Tangents to Curves★★★☆☆⏱ 5 min

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A tangent to a curve at a point is a straight line that touches the curve only at that point, with gradient equal to the derivative of the curve at that x-coordinate.

Calculate the derivative of the curve function to get the gradient function
Substitute the x-value of the given point into the derivative to get the gradient of the tangent
Use the point-slope form of a straight line $y - y_{1} = m (x - x_{1})$ where $(x_{1}, y_{1})$ is the point on the curve and $m$ is the tangent gradient

📐 Worked Example

Find the equation of the tangent to the curve $y = x^{3} - 2 x + 5$ at the point where $x = 2$ . Give your answer in the form $a x + b y + c = 0$ where a, b, c are integers.

1
First find the y-coordinate of the point when $x = 2$ :
2
$y = 2^3 - 2(2) +5 = 8 -4 +5 = 9. \text{ The point is } (2, 9)$
3
Differentiate the curve to get the gradient function:
4
$\frac{dy}{dx} = 3x^2 - 2$
5
Substitute $x = 2$ to find the tangent gradient:
6
$m = 3(2)^2 - 2 = 12 - 2 = 10$
7
Use point-slope form and rearrange to required form:
8
$y - 9 = 10(x - 2) \rightarrow 10x - y - 11 = 0$

Exam tip:

Always check you have the correct y-coordinate of the point first: many students forget to calculate this and use the x-value as y, losing easy marks.

4. Equations of Normals to Curves★★★★☆⏱ 5 min

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📘 Definition

Normal to a Curve

A straight line that passes through a point on a curve and is perpendicular to the tangent at that point.

Example:

If a tangent has gradient 3, the normal at the same point has gradient $- \frac{1}{3}$ .

The product of the gradients of two perpendicular lines is -1, so if the tangent gradient is $m_{t}$ , the normal gradient is $m_{n} = - \frac{1}{m _{t}}$ . The rest of the process for finding the normal equation is identical to finding the tangent equation.

📐 Worked Example

Find the equation of the normal to the curve $y = 2 x^{2} - \frac{1}{x}$ at the point where $x = 1$ . Give your answer in the form $y = m x + c$ .

1
Calculate the y-coordinate at $x = 1$ :
2
$y = 2(1)^2 - \frac{1}{1} = 2 - 1 = 1. \text{ The point is } (1, 1)$
3
Rewrite the function in index form and differentiate:
4
$y = 2x^2 - x^{-1} \rightarrow \frac{dy}{dx} = 4x + x^{-2}$
5
Find the tangent gradient at $x = 1$ :
6
$m_t = 4(1) + 1^{-2} = 4 +1 = 5$
7
Calculate the normal gradient, the negative reciprocal of $m_{t}$ :
8
$m_n = -\frac{1}{5}$
9
Use point-slope form and rearrange to required form:
10
$y -1 = -\frac{1}{5}(x -1) \rightarrow y = -\frac{1}{5}x + \frac{6}{5}$

Exam tip:

If the tangent gradient is 0 (horizontal tangent), the normal is a vertical line of the form $x = x_{1}$ , as it has an undefined gradient. Check for this case to avoid division by zero errors.

5. Common Pitfalls

Wrong move:

Trying to use product/quotient/chain rules to differentiate functions instead of expanding/simplifying first.

Why:

Edexcel P1 questions are designed to be solved without these rules, and using them leads to unnecessary errors, wasted time, and potential mark loss if applied incorrectly.

Correct move:

Always rewrite all terms as separate $a x^{n}$ terms (expand brackets, simplify indices, rewrite roots/denominator terms) before applying the power rule.

Wrong move:

Substituting the x-coordinate into the original curve function instead of the derivative to find the gradient of the tangent/normal.

Why:

The original function gives you the y-value of the point, not the gradient, so you will get the wrong line equation.

Correct move:

First differentiate the curve to get the gradient function, then substitute the x-value into the derivative to get the gradient of the tangent or normal.

Wrong move:

Using the tangent gradient directly for the normal equation instead of taking the negative reciprocal.

Why:

The normal is perpendicular to the tangent, so their gradients multiply to -1, they are not the same.

Correct move:

After finding the tangent gradient $m_{t}$ , calculate the normal gradient as $m_{n} = - \frac{1}{m _{t}}$ (unless $m_{t} = 0$ , in which case the normal is vertical $x = x_{1}$ ).

Wrong move:

Forgetting to reduce the power by 1 when applying the power rule, especially for negative or fractional powers.

Why:

This common arithmetic error leads to incorrect derivatives and all subsequent calculations being wrong.

Correct move:

Write down the exponent of each term clearly before differentiating, subtract 1 from the exponent, and double-check for negative/fractional exponents.

Wrong move:

Using the second derivative to classify stationary points in P1 exams.

Why:

Stationary points and their classification are explicitly out of scope for P1, so any work related to them will not gain marks and wastes time.

Correct move:

Only compute the second derivative if explicitly asked, and only explain its meaning as the rate of change of gradient; do not use it for stationary point analysis in P1.

6. Quick Reference Cheatsheet

Concept	Rule / Formula
Derivative meaning	Gradient of tangent to curve, instantaneous rate of change of $f (x)$
Power Rule	$\frac{d}{d x} (a x^{n}) = an x^{n - 1}$ for any rational $n$
Second derivative	Differentiate first derivative: $f^{''} (x) = \frac{d}{d x} (f^{'} (x))$
Tangent gradient	$m_{t} = f^{'} (x_{1})$ at point $(x_{1}, y_{1})$
Normal gradient	$m_{n} = - \frac{1}{m _{t}}$ (perpendicular to tangent)
Line equation	$y - y_{1} = m (x - x_{1})$ for point $(x_{1}, y_{1})$ and gradient $m$

7. Frequently Asked

Do I need to use chain/product/quotient rules for P1 differentiation?

No, per Edexcel P1 specifications, all functions can be expanded or simplified into a sum of $a x^{n}$ terms before differentiation. Advanced differentiation rules are tested in P2 and P3 only.

Can I use the second derivative to find stationary points in P1?

No, stationary points, their classification, and optimisation problems are explicitly out of scope for P1. You only need to compute $f^{''} (x)$ and explain its conceptual meaning as the rate of change of gradient.

Going deeper

syllabusEdexcel IAL Maths 2018 P1 SpecificationOfficial specification for P1 differentiation content

What's Next

Now that you have mastered P1 differentiation, you are ready to move to more advanced differentiation content in P2, including stationary points, the chain rule, and optimisation problems. These build directly on the power rule and line equation skills you have learned here, so make sure you can reliably differentiate sums of powers and find tangents/normals before progressing. You will also use differentiation skills in later applied units, including kinematics in Mechanics 1 and rate of change problems in Statistics 1.