Maths95 views·Updated 9 Sept 2026·24 pages

Mastering Differentiation in Higher Maths

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Differentiation is a fundamental calculus technique that tells you how...

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of 10

Index Rules Refresher

Before diving into differentiation, you'll need these index rules from N5 - they're absolutely essential for what's coming next. When multiplying powers with the same base, you add the indices: $x^a \times x^b = x^{a+b}$ . When dividing, you subtract them: $x^a \div x^b = x^{a-b}$ .

The trickier rules involve negative and fractional indices. Remember that $x^{-a} = \frac{1}{x^a}$ and $x^{a/b} = \sqrt[b]{x^a}$ . So $x^{1/2} = \sqrt{x}$ and $x^{3/2} = \sqrt{x^3}$ .

Watch out for this common mistake: $\frac{1}{5x} \neq 5x^{-1}$ . Instead, $\frac{1}{5x} = \frac{1}{5}x^{-1}$ or $(5x)^{-1}$ . Getting these basics right will make differentiation much smoother.

Quick tip: When dealing with fractional powers like $25^{3/2}$ , do the root first: $\sqrt{25^3} = 5^3 = 125$

of 10

Introduction to Differentiation

Straight lines have a constant gradient, but curves are more interesting - their steepness changes at every point. To find the gradient at any specific point on a curve, we use the tangent line at that point.

Differentiation is the process that finds this gradient for us. The basic rule is surprisingly simple: if $y = x^n$ , then $\frac{dy}{dx} = nx^{n-1}$ . The symbol $\frac{dy}{dx}$ represents the derivative - it's just notation for "the gradient of y with respect to x".

This process might seem like magic at first, but applying it is straightforward. The mathematical proof behind why it works is complex, but you don't need to worry about that right now.

Remember: The derivative tells you the gradient of the tangent at any point on a curve

of 10

Basic Differentiation Examples

Let's put the power rule into action with some examples. For $y = x^2$ , we get $\frac{dy}{dx} = 2x$ . For $y = 3x^3$ , we get $\frac{dy}{dx} = 9x^2$ (multiply by the power, then reduce the power by 1).

The rule works brilliantly with multiple terms too. For $y = 5x^2 + 4x$ , you differentiate each term separately: $\frac{dy}{dx} = 10x + 4$ . Constants disappear when differentiated, so $y = 4x^2 - 2x + 1$ becomes $\frac{dy}{dx} = 8x - 2$ .

Negative and fractional powers follow the same rule. For $y = \sqrt{x} = x^{1/2}$ , we get $\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$ . For $y = 3x^{-2}$ , we get $\frac{dy}{dx} = -6x^{-3} = -\frac{6}{x^3}$ .

Pro tip: Always convert your final answer to positive indices - it looks much neater!

of 10

Notation and Preparation

You'll see derivatives written in two main ways: $\frac{dy}{dx}$ when $y =$ something, or $f'(x)$ (pronounced "f dash x") when $f(x) =$ something. Both mean exactly the same thing - the derivative of the function.

Before differentiating, you often need to prepare your function by converting roots and fractions into index form. For example, $\sqrt[3]{x^2}$ becomes $x^{2/3}$ , and $\frac{4}{x^2}$ becomes $4x^{-2}$ .

Once prepared, differentiate using the power rule as normal. So $y = x^{2/3}$ gives $\frac{dy}{dx} = \frac{2}{3}x^{-1/3}$ , and $f(x) = 4x^{-2}$ gives $f'(x) = -8x^{-3}$ .

Key insight: Converting everything to index form first makes differentiation much more systematic

of 10

Expanding and Simplifying Before Differentiation

When faced with brackets or fractions, expand or simplify first before differentiating. For $y = x^2(x-2)$ , multiply out to get $y = x^3 - 2x^2$ , then differentiate to get $\frac{dy}{dx} = 3x^2 - 4x$ .

With more complex expressions like $f(x) = \sqrt{x}(x - 2x^2)$ , convert the square root first: $f(x) = x^{1/2}(x - 2x^2) = x^{3/2} - 2x^{5/2}$ . Then differentiate: $f'(x) = \frac{3}{2}x^{1/2} - 5x^{3/2}$ .

Algebraic fractions work similarly. For $f(x) = \frac{2x^3 - 5x^4}{x^2}$ , divide each term by $x^2$ to get $f(x) = 2x - 5x^2$ . Then differentiate normally: $f'(x) = 2 - 10x$ .

Strategy: Always simplify first - it makes the differentiation much easier and reduces errors

of 10

The Mathematical Definition

The derivative is actually the limit of the gradient between two points as they get infinitely close together. This is written as $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ , but you don't need to use this formula directly.

When we differentiate functions, we're finding the gradient of the tangent to the curve at any given point. This gradient also represents the rate of change of the function at that point.

For example, with $f(x) = 4x^3$ , we get $f'(x) = 12x^2$ . To find the gradient at $x = -2$ , we substitute: $f'(-2) = 12 \times (-2)^2 = 48$ . So the tangent has a gradient of 48 at that point.

Real-world connection: Rate of change appears everywhere - speed is the rate of change of distance, acceleration is the rate of change of speed

of 10

Finding Gradients at Specific Points

To find the gradient at a specific point, differentiate the function then substitute the x-value. For $y = 4\sqrt{x}$ at $x = 9$ , first rewrite as $y = 4x^{1/2}$ , then differentiate to get $\frac{dy}{dx} = 2x^{-1/2} = \frac{2}{\sqrt{x}}$ .

Substituting $x = 9$ : $\frac{dy}{dx} = \frac{2}{\sqrt{9}} = \frac{2}{3}$ . So the gradient of the tangent at $x = 9$ is $\frac{2}{3}$ .

For more complex functions like $f(x) = 3x^3 + 3x^2 - 2x$ , differentiate to get $f'(x) = 9x^2 + 6x - 2$ . At $x = 3$ : $f'(3) = 9(9) + 6(3) - 2 = 81 + 18 - 2 = 97$ . This means the function is changing very rapidly at this point.

Remember: The derivative gives you the gradient; substituting a value gives you the gradient at that specific point

of 10

Equations of Tangent Lines

Since a tangent is a straight line, you need a point and a gradient to find its equation using $y - b = m(x - a)$ . The derivative gives you the gradient, and you find the point by substituting into the original function.

For the curve $y = x^2 - 7x + 10$ at $x = 4$ : first find the point by substituting $x = 4$ into the original equation: $y = 16 - 28 + 10 = -2$ . So the point is $(4, -2)$ .

Next, find the gradient by differentiating: $\frac{dy}{dx} = 2x - 7$ . At $x = 4$ : gradient = $2(4) - 7 = 1$ . Using the point-slope form: $y - (-2) = 1(x - 4)$ , which simplifies to $y = x - 6$ .

Method: Find the point (substitute x into original function), find the gradient (substitute x into derivative), then use point-slope form

of 10

Increasing and Decreasing Functions

The derivative tells you whether a function is going up or down. If $f'(x) > 0$ , the function is strictly increasing (going uphill). If $f'(x) < 0$ , it's strictly decreasing (going downhill). When $f'(x) = 0$ , the function is stationary (flat).

Think of it like walking on a hill - positive gradient means you're walking uphill, negative means downhill, and zero means you're on flat ground or at a peak/valley.

These concepts are crucial for understanding the shape and behaviour of curves. You'll use them constantly when sketching graphs and solving optimisation problems.

Visual tip: Imagine the curve as a roller coaster - the derivative tells you whether you're going up, down, or momentarily flat

of 10

Differentiation - Higher Maths – page 10

Determining Intervals of Increase and Decrease

To show a function is strictly increasing at a point, prove that $f'(x) > 0$ there. For $p(x) = x^3 - 4x^2 + 3$ at $x = 3$ : $p'(x) = 3x^2 - 8x$ , so $p'(3) = 27 - 24 = 3 > 0$ . Since the derivative is positive, the function is increasing.

To find where a function is decreasing, solve $f'(x) < 0$ . For $f(x) = x^3 + 6x^2 - 2$ : $f'(x) = 3x^2 + 12x = 3x(x + 4)$ . This is negative when $-4 < x < 0$ .

Some functions are always increasing. For $f(x) = x^3 + 3x^2 + 5x + 2$ : $f'(x) = 3x^2 + 6x + 5$ . Completing the square: $f'(x) = 3(x + 1)^2 + 2$ . Since this is always positive, the function always increases.

Technique: Use completing the square to prove a quadratic expression is always positive or negative

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Maths95 views·Updated 9 Sept 2026·24 pages

Mastering Differentiation in Higher Maths

eva@eva_lbjt3

Add to Google sources

Differentiation is a fundamental calculus technique that tells you how quickly a function is changing at any point. Think of it like finding the steepness of a hill at different spots - sometimes it's gentle, sometimes steep, and sometimes you're...

of 10

Index Rules Refresher

The trickier rules involve negative and fractional indices. Remember that $x^{-a} = \frac{1}{x^a}$ and $x^{a/b} = \sqrt[b]{x^a}$ . So $x^{1/2} = \sqrt{x}$ and $x^{3/2} = \sqrt{x^3}$ .

Watch out for this common mistake: $\frac{1}{5x} \neq 5x^{-1}$ . Instead, $\frac{1}{5x} = \frac{1}{5}x^{-1}$ or $(5x)^{-1}$ . Getting these basics right will make differentiation much smoother.

Quick tip: When dealing with fractional powers like $25^{3/2}$ , do the root first: $\sqrt{25^3} = 5^3 = 125$

of 10

Introduction to Differentiation

This process might seem like magic at first, but applying it is straightforward. The mathematical proof behind why it works is complex, but you don't need to worry about that right now.

Remember: The derivative tells you the gradient of the tangent at any point on a curve

of 10

Basic Differentiation Examples

Pro tip: Always convert your final answer to positive indices - it looks much neater!

of 10

Notation and Preparation

Once prepared, differentiate using the power rule as normal. So $y = x^{2/3}$ gives $\frac{dy}{dx} = \frac{2}{3}x^{-1/3}$ , and $f(x) = 4x^{-2}$ gives $f'(x) = -8x^{-3}$ .

Key insight: Converting everything to index form first makes differentiation much more systematic

of 10

Expanding and Simplifying Before Differentiation

Algebraic fractions work similarly. For $f(x) = \frac{2x^3 - 5x^4}{x^2}$ , divide each term by $x^2$ to get $f(x) = 2x - 5x^2$ . Then differentiate normally: $f'(x) = 2 - 10x$ .

Strategy: Always simplify first - it makes the differentiation much easier and reduces errors

of 10

The Mathematical Definition

When we differentiate functions, we're finding the gradient of the tangent to the curve at any given point. This gradient also represents the rate of change of the function at that point.

For example, with $f(x) = 4x^3$ , we get $f'(x) = 12x^2$ . To find the gradient at $x = -2$ , we substitute: $f'(-2) = 12 \times (-2)^2 = 48$ . So the tangent has a gradient of 48 at that point.

Real-world connection: Rate of change appears everywhere - speed is the rate of change of distance, acceleration is the rate of change of speed

of 10

Finding Gradients at Specific Points

Substituting $x = 9$ : $\frac{dy}{dx} = \frac{2}{\sqrt{9}} = \frac{2}{3}$ . So the gradient of the tangent at $x = 9$ is $\frac{2}{3}$ .

Remember: The derivative gives you the gradient; substituting a value gives you the gradient at that specific point

of 10

Equations of Tangent Lines

For the curve $y = x^2 - 7x + 10$ at $x = 4$ : first find the point by substituting $x = 4$ into the original equation: $y = 16 - 28 + 10 = -2$ . So the point is $(4, -2)$ .

Next, find the gradient by differentiating: $\frac{dy}{dx} = 2x - 7$ . At $x = 4$ : gradient = $2(4) - 7 = 1$ . Using the point-slope form: $y - (-2) = 1(x - 4)$ , which simplifies to $y = x - 6$ .

Method: Find the point (substitute x into original function), find the gradient (substitute x into derivative), then use point-slope form

of 10

Increasing and Decreasing Functions

Think of it like walking on a hill - positive gradient means you're walking uphill, negative means downhill, and zero means you're on flat ground or at a peak/valley.

These concepts are crucial for understanding the shape and behaviour of curves. You'll use them constantly when sketching graphs and solving optimisation problems.

Visual tip: Imagine the curve as a roller coaster - the derivative tells you whether you're going up, down, or momentarily flat

of 10

Determining Intervals of Increase and Decrease

To find where a function is decreasing, solve $f'(x) < 0$ . For $f(x) = x^3 + 6x^2 - 2$ : $f'(x) = 3x^2 + 12x = 3x(x + 4)$ . This is negative when $-4 < x < 0$ .

Technique: Use completing the square to prove a quadratic expression is always positive or negative

We thought you’d never ask...

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That's right! Enjoy free access to study content, connect with fellow students, and get instant help – all at your fingertips.

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Index Rules Refresher

Introduction to Differentiation

Basic Differentiation Examples

Notation and Preparation

Expanding and Simplifying Before Differentiation

The Mathematical Definition

Finding Gradients at Specific Points

Equations of Tangent Lines

Increasing and Decreasing Functions

Determining Intervals of Increase and Decrease

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Finding Gradients at Specific Points

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