Differentiation

Q: What does the derivative $\frac{dy}{dx}$ represent geometrically on a curve?

It represents the gradient (slope) of the tangent to the curve at any specific point $(x, y)$.

Q: What is the Power Rule for differentiating a term of the form $ax^n$?

If $y = ax^n$, then the derivative is $\frac{dy}{dx} = anx^{n-1}$.

Q: How must you rewrite terms like $\frac{1}{x^n}$ or $\sqrt{x}$ before differentiating?

You must rewrite them in index form : $\frac{1}{x^n} = x^{-n}$ and $\sqrt{x} = x^{1/2}$.

Q: What is the derivative of a constant term (e.g., $y = 7$)?

The derivative of a constant is always zero: $\frac{d}{dx}(c) = 0$.

Q: What is the derivative of the natural log function $y = \ln(kx)$?

The derivative is $\frac{dy}{dx} = \frac{1}{x}$, regardless of the constant $k$.

Q: What is the Chain Rule for a function of a function?

The rule is $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$.

Q: How do you find the gradient of a normal to a curve at a specific point?

Find the tangent gradient ($m_T$) using differentiation, then calculate the negative reciprocal : $m_N = -\frac{1}{m_T}$.

Q: What condition is true at any stationary point on a curve?

The gradient is zero: $\frac{dy}{dx} = 0$.

Q: How does the second derivative $\frac{d^2y}{dx^2}$ help classify a stationary point?

If $\frac{d^2y}{dx^2} > 0$ the point is a minimum ; if \(\frac{d^2y}{dx^2} maximum .

Practice

Pure Mathematics (YPM01) · Pearson Edexcel International A Level · 8 min read · Updated: 6 Sept 2026

Welcome to the World of Differentiation!

Hello Mathematician! Differentiation is one of the most powerful tools in Pure Mathematics. Don't worry if the term sounds intimidating—it's simply a method used to measure how fast things are changing.

In this chapter, you will learn how to find the instantaneous rate of change of a function, which is essentially finding the exact steepness (or gradient) of a curve at a single point. This skill is vital for solving real-world problems involving speed, growth, and optimization.

1. Understanding the Concept: Gradient and Rate of Change

1.1 What is Differentiation?

Imagine driving a car. If you measure the distance covered over a whole hour, you get your average speed. But what about the speed shown on your speedometer right now, this exact second? That is the instantaneous rate of change.

In math, the instantaneous rate of change is called the Derivative.
We often use the notation $\frac{dy}{dx}$, which means "the change in $y$ divided by the change in $x$" at an infinitely small interval.
Geometrically, $\frac{dy}{dx}$ represents the gradient of the tangent to the curve $y = f(x)$ at any point $(x, y)$.

1.2 Finding the Gradient Function

Instead of finding the gradient at just one point, differentiation allows us to find a new function—the gradient function—which gives us the gradient for every value of $x$.

Key Notation Review:
If the function is $y = f(x)$:
The first derivative (gradient function) is written as:

$\frac{dy}{dx}$
$f'(x)$

Key Takeaway: Differentiation finds the exact gradient of a curve at any specific point, telling us the instantaneous rate of change.

2. The Power Rule: Differentiating Polynomials

The Power Rule is your bread and butter in calculus. It allows you to quickly differentiate terms of the form $ax^n$.

2.1 The Core Rule

If $y = ax^n$, where $a$ and $n$ are constants, then the derivative is:

$$ \frac{dy}{dx} = anx^{n-1} $$

Step-by-Step Method:

Multiply: Multiply the existing coefficient ($a$) by the power ($n$).
Reduce: Reduce the power ($n$) by one.

Memory Aid: "Multiply down, power down."

Example:
If $y = 4x^3$:
1. Multiply: $4 \times 3 = 12$
2. Reduce: $3 - 1 = 2$
So, $\frac{dy}{dx} = 12x^2$.

2.2 Differentiating Complex Forms (Indices Review)

Before applying the power rule, you MUST rewrite the function so that all terms are in the form $ax^n$. This requires knowledge of indices.

Original Form ($y$)	Index Form (Ready to Differentiate)
$y = \frac{1}{x^2}$	$y = x^{-2}$
$y = \sqrt{x}$	$y = x^{1/2}$
$y = 5\sqrt{x^3}$	$y = 5x^{3/2}$

Common Mistake Alert!

If $y = \frac{3}{x}$, many students differentiate this as if it were a quotient. STOP! Rewrite it first: $y = 3x^{-1}$.
Then, $\frac{dy}{dx} = (3)(-1)x^{-1-1} = -3x^{-2} = -\frac{3}{x^2}$.

2.3 Differentiating Sums, Differences, and Constants

If a function is a sum or difference of terms, you simply differentiate each term separately.

Constants: If $y = c$ (where $c$ is a constant, e.g., $y=7$), the gradient is zero because a horizontal line has no steepness. $$ \frac{d}{dx} (c) = 0 $$
Linear Terms: If $y = ax$ (e.g., $y=5x$), this is really $y = 5x^1$. $$ \frac{d}{dx} (5x) = (5)(1)x^{1-1} = 5x^0 = 5 $$

Example of a combined function:
If $y = 3x^4 - 2x + 7$
$$ \frac{dy}{dx} = 12x^3 - 2 + 0 = 12x^3 - 2 $$

Key Takeaway: Rewrite everything as $ax^n$, then use the Power Rule: Multiply by the power, reduce the power by one. Constants disappear!

3. Differentiating Exponential and Logarithmic Functions (P2 Specials)

In P2, you are expected to differentiate functions involving the natural exponential $e^x$ and the natural logarithm $\ln x$.

3.1 The Exponential Function ($e^x$)

The exponential function is unique because its derivative is itself!

If $y = e^x$, then $\frac{dy}{dx} = e^x$.

When the power is a simple multiple of $x$:

$$ \text{If } y = e^{kx}, \text{ then } \frac{dy}{dx} = ke^{kx} $$

Trick: You just multiply the function by the coefficient of $x$.

Example:
1. If $y = e^{5x}$, then $\frac{dy}{dx} = 5e^{5x}$.
2. If $y = 3e^{-2x}$, then $\frac{dy}{dx} = 3 \times (-2)e^{-2x} = -6e^{-2x}$.

3.2 The Logarithmic Function ($\ln x$)

The derivative of the natural log function is surprisingly simple:

$$ \text{If } y = \ln x, \text{ then } \frac{dy}{dx} = \frac{1}{x} $$

What if there is a coefficient inside the logarithm, like $y = \ln(kx)$?

$$ \text{If } y = \ln(kx), \text{ then } \frac{dy}{dx} = \frac{1}{x} $$

Did you know? Because of the laws of logs ($\ln(kx) = \ln k + \ln x$), and since $\ln k$ is just a constant, when you differentiate $\ln(kx)$, the $\ln k$ term differentiates to zero, leaving only $\frac{1}{x}$!

Example:
If $y = 4 \ln(3x) + x^2$, then $\frac{dy}{dx} = 4 \left( \frac{1}{x} \right) + 2x = \frac{4}{x} + 2x$.

Key Takeaway: $e^x$ differentiates to itself (times the coefficient of x). $\ln(x)$ or $\ln(kx)$ differentiates to $\frac{1}{x}$.

4. The Chain Rule (Function of a Function)

The Chain Rule is used when you have a function nested inside another function—a "function of a function." Think of it like a set of Russian dolls or peeling an onion. You must differentiate the outer layer first, then the inner layer.

In P2, you typically encounter the Chain Rule when differentiating brackets raised to a power, or when the power of $e$ is more complex than just $kx$.

4.1 The Formal Rule (Simplified)

If $y$ depends on $u$, and $u$ depends on $x$, we write: $$ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} $$

4.2 The Intuitive Step-by-Step Method

Let $y = (f(x))^n$.

Differentiate the Outside: Apply the Power Rule to the whole bracket (treat the content inside the bracket as a single variable).
Leave the Inside Alone: Do NOT change the expression inside the bracket yet.
Multiply by the Derivative of the Inside: Differentiate the expression inside the bracket, and multiply your result by it.

Example 1: Polynomial Bracket
Differentiate $y = (2x^3 - 5)^4$.
1. Outside: $4(2x^3 - 5)^3$
2. Derivative of Inside ($2x^3 - 5$): $6x^2$
3. Combine: $\frac{dy}{dx} = 4(2x^3 - 5)^3 \times (6x^2)$
4. Simplify: $\frac{dy}{dx} = 24x^2(2x^3 - 5)^3$

Example 2: Exponential Chain
Differentiate $y = e^{x^2 + 3x}$.
1. Outside: $e^{x^2 + 3x}$ (The derivative of $e^{\text{something}}$ is $e^{\text{something}}$).
2. Derivative of Inside ($x^2 + 3x$): $2x + 3$
3. Combine: $\frac{dy}{dx} = e^{x^2 + 3x} \times (2x + 3)$
4. Simplify: $\frac{dy}{dx} = (2x + 3)e^{x^2 + 3x}$

Key Takeaway: When using the Chain Rule, always multiply the result by the derivative of the "inner" function.

5. Applications of Differentiation

5.1 Finding Equations of Tangents and Normals

Since $\frac{dy}{dx}$ gives the gradient of the curve, we can use it to find the equations of lines that interact with the curve at a specific point $(x_1, y_1)$. We use the straight-line equation: $y - y_1 = m(x - x_1)$.

Steps to find the Equation of a Tangent:

Find the $y$-coordinate for the given $x$-coordinate (if not provided).
Differentiate the function, $y$, to find the gradient function, $\frac{dy}{dx}$.
Substitute the $x$-coordinate into $\frac{dy}{dx}$ to find the gradient of the tangent, $m_T$.
Use the point $(x_1, y_1)$ and gradient $m_T$ in the equation $y - y_1 = m_T(x - x_1)$.

The Normal Line

The Normal is the line perpendicular (at 90°) to the tangent at that point.

If the gradient of the tangent is $m_T$, the gradient of the normal, $m_N$, is the negative reciprocal: $$ m_N = -\frac{1}{m_T} $$

Example: If the tangent gradient is 4, the normal gradient is $-\frac{1}{4}$. If the tangent gradient is $-\frac{2}{3}$, the normal gradient is $\frac{3}{2}$.

5.2 Stationary Points (Maxima and Minima)

A Stationary Point (also called a turning point) is a point on a curve where the gradient is exactly zero.

$$ \text{At a stationary point: } \frac{dy}{dx} = 0 $$

There are three types of stationary points: Maximum, Minimum, and Point of Inflection (which is rarely tested in P2).

Determining the Nature (Max or Min): The Second Derivative Test

To determine if a stationary point is a maximum or minimum, we use the Second Derivative, denoted by $\frac{d^2y}{dx^2}$ or $f''(x)$. This is simply the derivative of the first derivative.

Step-by-Step for Finding Stationary Points:

Find $\frac{dy}{dx}$: Differentiate the original function.
Set to Zero: Solve $\frac{dy}{dx} = 0$ to find the $x$-coordinates of the stationary points.
Find $\frac{d^2y}{dx^2}$: Differentiate $\frac{dy}{dx}$ to find the second derivative.
Test the Points: Substitute each $x$-value found in step 2 into $\frac{d^2y}{dx^2}$:
- If $\frac{d^2y}{dx^2} > 0$, it is a Minimum Point (the curve is curving upwards, like a smiling face).
- If $\frac{d^2y}{dx^2} < 0$, it is a Maximum Point (the curve is curving downwards, like a sad face).
Find $y$: Substitute the $x$-values back into the original equation $y = f(x)$ to find the $y$-coordinates.

Encouragement: Don't panic about remembering which sign means which point! Think of it this way: Positive curvature (++) is happy, so it's a MINIMUM. Negative curvature (--) is sad, so it's a MAXIMUM.

Key Takeaway: The gradient $\frac{dy}{dx}$ gives you the tangent slope. Setting it to zero finds turning points. The second derivative $\frac{d^2y}{dx^2}$ determines if those points are maxima or minima.

Quick Review Checklist

Before moving on, make sure you can confidently perform these P2 differentiation skills:

Rewrite roots and fractions as $x^n$ form.
Use the power rule for terms like $3x^{-4}$.
Differentiate $e^{kx}$ and $\ln(kx)$.
Apply the Chain Rule to simple polynomials and exponentials (e.g., $(4x+1)^6$ and $e^{x^2}$).
Find the gradient of a tangent at a specific point.
Find the gradient of a normal using the negative reciprocal.
Solve $\frac{dy}{dx} = 0$ to find stationary points.
Use $\frac{d^2y}{dx^2}$ to classify stationary points.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What does the derivative $\frac{dy}{dx}$ represent geometrically on a curve?

It represents the gradient (slope) of the tangent to the curve at any specific point $(x, y)$.

What is the Power Rule for differentiating a term of the form $ax^n$?

If $y = ax^n$, then the derivative is $\frac{dy}{dx} = anx^{n-1}$.

How must you rewrite terms like $\frac{1}{x^n}$ or $\sqrt{x}$ before differentiating?

You must rewrite them in index form: $\frac{1}{x^n} = x^{-n}$ and $\sqrt{x} = x^{1/2}$.

What is the derivative of a constant term (e.g., $y = 7$)?

The derivative of a constant is always zero: $\frac{d}{dx}(c) = 0$.

How do you differentiate the natural exponential function $y = e^{kx}$?

$\frac{dy}{dx} = ke^{kx}$.

What is the derivative of the natural log function $y = \ln(kx)$?

The derivative is $\frac{dy}{dx} = \frac{1}{x}$, regardless of the constant $k$.

What is the Chain Rule for a function of a function?

The rule is $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$.

How do you find the gradient of a normal to a curve at a specific point?

Find the tangent gradient ($m_T$) using differentiation, then calculate the negative reciprocal: $m_N = -\frac{1}{m_T}$.

What condition is true at any stationary point on a curve?

The gradient is zero: $\frac{dy}{dx} = 0$.

How does the second derivative $\frac{d^2y}{dx^2}$ help classify a stationary point?

If $\frac{d^2y}{dx^2} > 0$ the point is a minimum; if $\frac{d^2y}{dx^2} < 0$ it is a maximum.

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Original Form (\(y\))	Index Form (Ready to Differentiate)
\(y = \frac{1}{x^2}\)	\(y = x^{-2}\)
\(y = \sqrt{x}\)	\(y = x^{1/2}\)
\(y = 5\sqrt{x^3}\)	\(y = 5x^{3/2}\)