Differentiation using standard results Revision Notes for AQA GCSE Further Maths

Differentiation using standard results

Understanding differentiation notation

When working with calculus, you'll encounter different ways to write the same mathematical concepts. If you have a function like $y = 2x^2 + x - 3$ , there are two main ways to express this. You can write it using function notation as $f(x) = 2x^2 + x - 3$ , which you would read aloud as "f of x equals two x squared plus x minus three."

Note

The derivative of a function can be written in two equivalent ways. Using function notation, we write $f'(x)$ , which is pronounced "f dashed x". Alternatively, we can write $\frac{dy}{dx}$ , which represents "the derivative of y with respect to x." Both expressions mean exactly the same thing - they tell us how quickly the function is changing at any given point.

The fundamental power rule

The most important rule for differentiation is the power rule, which works for any function of the form $y = kx^n$ , where $k$ is a constant and $n$ is any real number.

Important

The Power Rule: When you have $y = kx^n$ , the derivative is:

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

Memory aid: Multiply by the power of x, then reduce the power by 1

Let's see how this works with some basic examples:

For $y = x^2$ , we multiply by the power (2) and reduce the power by 1: $\frac{dy}{dx} = 2x^1 = 2x$
For $y = x^3$ , we get: $\frac{dy}{dx} = 3x^2$
For $y = 4x^3$ , we get: $\frac{dy}{dx} = 12x^2$

The rule also applies when the power is zero or one:

For $y = x^1$ (which is just $y = x$ ), we get: $\frac{dy}{dx} = 1x^0 = 1$
For $y = x^0$ (which equals 1, a constant), we get: $\frac{dy}{dx} = 0$

This last result reveals an important principle: the derivative of any constant is always zero.

Working with negative powers and fractions

Many functions contain fractions, but these can be handled easily by converting them to negative powers first. This conversion is essential because it allows us to apply the power rule consistently.

Here's how the conversion works:

$\frac{1}{x}$ becomes $x^{-1}$
$\frac{1}{x^2}$ becomes $x^{-2}$
$\frac{3}{x^2}$ becomes $3x^{-2}$
$\frac{2}{3x^2}$ becomes $\frac{2}{3}x^{-2}$

Once you've made this conversion, you can apply the power rule exactly as before.

Example

Worked Example: Differentiating with Negative Powers

Example 1: Differentiate $y = x^{-3}$ Using the power rule: $\frac{dy}{dx} = -3x^{-4}$

Example 2: Differentiate $y = \frac{3}{x^2}$ Step 1: Rewrite as $y = 3x^{-2}$ Step 2: Apply power rule: $\frac{dy}{dx} = 3(-2)x^{-3} = -6x^{-3} = -\frac{6}{x^3}$

Example 3: Differentiate $y = 2x^{-4}$
Using the power rule: $\frac{dy}{dx} = -8x^{-5}$

The key insight here is that you must leave fractions in their negative power form until the very end of your calculation, when you can convert back to fraction form if needed.

Differentiating sums and differences of functions

One of the most useful properties of differentiation is that you can handle sums and differences by treating each term separately. When you have a function that consists of multiple terms added or subtracted together, you simply differentiate each term individually and then combine the results.

For example, if you have a function like $y = 4x^3 + 3x$ , you can think of this as the sum of two simpler functions: $4x^3$ and $3x$ . To differentiate the entire expression, you differentiate each part separately:

The derivative of $4x^3$ is $12x^2$
The derivative of $3x$ is $3$
Therefore, $\frac{dy}{dx} = 12x^2 + 3$

Note

This principle works regardless of how many terms your function contains, and it applies to both addition and subtraction. Each term can be differentiated independently, making complex expressions much more manageable.

Example

Worked Example: Differentiating Sums with Fractions

Differentiate $y = \frac{x^2}{2} - \frac{2}{3x^2}$

Step 1: Rewrite in standard form: $y = \frac{1}{2}x^2 - \frac{2}{3}x^{-2}$

Step 2: Differentiate each term:

$\frac{d}{dx}\left\[\frac{1}{2}x^2\right] = \frac{1}{2}(2x) = x$
$\frac{d}{dx}\left\[-\frac{2}{3}x^{-2}\right] = -\frac{2}{3}(-2)x^{-3} = \frac{4}{3}x^{-3}$

Step 3: Combine results: $\frac{dy}{dx} = x + \frac{4}{3}x^{-3} = x + \frac{4}{3x^3}$

Finding gradients and rates of change

Once you've found the derivative of a function, you can use it to find the gradient at any specific point on the curve. This is done by substituting the x-coordinate of the point into the derivative expression.

The derivative also tells us the rate of change of y with respect to x. This is exactly the same as the gradient - it's just a different way of describing the same mathematical concept.

Example

Worked Example: Finding Gradient at a Point

Given $y = 2x^3 - 3x + 4$ , find the gradient at the point $(2, 14)$

Step 1: Find the derivative: $\frac{dy}{dx} = 6x^2 - 3$

Step 2: Substitute $x = 2$ : $\frac{dy}{dx} = 6(2)^2 - 3 = 6(4) - 3 = 24 - 3 = 21$

Therefore, the gradient of the curve at the point $(2, 14)$ is 21.

Expressions requiring expansion or division

Sometimes you'll encounter expressions that don't immediately look like they can be differentiated using the power rule. In these cases, you need to manipulate the expression first by expanding brackets or dividing fractions.

The key principle here is to always simplify the expression into a sum or difference of power terms before attempting to differentiate.

Example

Worked Example: Simplifying Before Differentiating

Example 1: Differentiate $y = x^2(x^2 - 4)$

Step 1: Expand the brackets first: $y = x^4 - 4x^2$ Step 2: Differentiate: $\frac{dy}{dx} = 4x^3 - 8x$

Example 2: Differentiate $y = \frac{x^3 + x^2}{x}$

Step 1: Divide each term by x: $y = x^2 + x$
Step 2: Differentiate: $\frac{dy}{dx} = 2x + 1$

Comprehensive worked example

Example

Worked Example: Complete Differentiation Problem

Given: $y = 5x^2 - \frac{5}{x^2} + 2$

Find: (i) $\frac{dy}{dx}$ , (ii) the gradient at point $(1, 2)$ , (iii) the rate of change when $x = -1$

Solution:

(i) Finding the derivative: Step 1: Rewrite in standard form: $y = 5x^2 - 5x^{-2} + 2$

Step 2: Differentiate each term:

$\frac{d}{dx}\[5x^2] = 10x$
$\frac{d}{dx}\[-5x^{-2}] = -5(-2)x^{-3} = 10x^{-3} = \frac{10}{x^3}$
$\frac{d}{dx}\[2] = 0$

Therefore: $\frac{dy}{dx} = 10x + \frac{10}{x^3}$

(ii) Gradient at point $(1, 2)$ : Substitute $x = 1$ : $\frac{dy}{dx} = 10(1) + \frac{10}{(1)^3} = 10 + 10 = 20$

(iii) Rate of change when $x = -1$ : Substitute $x = -1$ : $\frac{dy}{dx} = 10(-1) + \frac{10}{(-1)^3} = -10 + \frac{10}{-1} = -10 - 10 = -20$

Common exam tips and traps

Important

Essential Tips for Success:

Tip 1: Always convert fractions to negative powers before differentiating. This avoids confusion and makes the power rule applicable.

Tip 2: When dealing with constants, remember that they differentiate to zero. Many students forget this step.

Tip 3: For complex expressions, always simplify first. Look for opportunities to expand brackets or cancel terms.

Tip 4: Double-check your arithmetic, especially with negative powers and signs. These are common sources of errors.

Common Traps to Avoid:

Don't try to differentiate fractions in their original form - always convert to negative powers first
Remember that $x^{-1}$ becomes $-1x^{-2}$ , not $\frac{x^{-2}}{-1}$ . The coefficient matters
When substituting values to find gradients, be extra careful with negative numbers and their powers

Key takeaways

Summary

Key Points to Remember:

The power rule is fundamental: For $y = kx^n$ , we get $\frac{dy}{dx} = nkx^{n-1}$ . Always multiply by the power first, then reduce the power by 1
Convert fractions to negative powers: Write $\frac{1}{x^2}$ as $x^{-2}$ before differentiating to make the power rule applicable
Treat each term separately: When differentiating sums or differences, handle each term individually and combine the results
Simplify complex expressions first: Expand brackets or divide fractions to create simple power terms before differentiating
The derivative gives both gradient and rate of change: These are the same concept expressed in different ways - substitute x-values into $\frac{dy}{dx}$ to find either one

Differentiation using standard results (AQA GCSE Further Maths): Revision Notes