# Differentiation

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Differentiation

We use differentiation to find the gradient of a graph at any given point – that’s the steepness of the graph.

A Level   ## Notation and Formula

So, let’s say we’ve got a function $f(\textcolor{blue}{x}) = \textcolor{limegreen}{y} = \textcolor{blue}{x}^\textcolor{red}{n}$.

Our derivative, $f'(x)$ or $\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}$, is the result of differentiating $\textcolor{limegreen}{y}$ with respect to $\textcolor{blue}{x}$, given by the equation

$f'(\textcolor{blue}{x}) = \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \textcolor{red}{n}\textcolor{blue}{x}^{\textcolor{red}{n}-1}$

This derivative gives a formula for the graph’s gradient for any value of $\textcolor{blue}{x}$.

As an example, let $\textcolor{limegreen}{y} = \textcolor{blue}{x}^\textcolor{red}{2}$.

Then

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \textcolor{red}{2}\textcolor{blue}{x}^{\textcolor{red}{2} - 1} = \textcolor{red}{2}\textcolor{blue}{x}$

A Level   ## Negative and Fractional Roots

Fear not, we’ll just apply the same formula as before.

Here’s a couple of examples:

Let $\textcolor{limegreen}{y} = \dfrac{1}{\textcolor{blue}{x}} = \textcolor{blue}{x}^{\textcolor{red}{-1}}$. Then

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \textcolor{red}{-1}\textcolor{blue}{x}^{\textcolor{red}{-1} - 1} = - \textcolor{blue}{x}^{-2} = \dfrac{-1}{\textcolor{blue}{x}^2}$

Let $\textcolor{limegreen}{y} = \sqrt{\textcolor{blue}{x}} = \textcolor{blue}{x}^{\textcolor{red}{\frac{1}{2}}}$. Then

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}= \textcolor{red}{\dfrac{1}{2}}\textcolor{blue}{x}^{\textcolor{red}{\frac{1}{2}} - 1} = \textcolor{red}{\dfrac{1}{2}}\textcolor{blue}{x}^{\frac{-1}{2}} = \dfrac{1}{2\sqrt{\textcolor{blue}{x}}}$

A Level   ## Differentiating a Linear Combination of Terms

For functions which involve multiple terms, we should differentiate each term individually.

So, say $\textcolor{limegreen}{y} = \textcolor{blue}{x}^\textcolor{red}{2} + \textcolor{blue}{x} - 1$.

Then

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \textcolor{red}{2}\textcolor{blue}{x}^{\textcolor{red}{2} - 1} + \textcolor{red}{1}\textcolor{blue}{x}^{\textcolor{red}{1} - 1} - (1 \times \textcolor{red}{0})\textcolor{blue}{x}^{\textcolor{red}{0} - 1}$

$= 2\textcolor{blue}{x} + 1 - 0$

$= 2x + 1$

Notice how the term without $\textcolor{blue}{x}$, “$- 1$” becomes $0$ in the expression for $\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}$. This is true for all constant terms (i.e. those which do not have an $\textcolor{blue}{x}$ term).

A Level   A Level   ## Example: Simplification Before Differentiation

Say we have a graph of $\textcolor{limegreen}{y} = (\textcolor{blue}{x}^\textcolor{red}{2} + 1)(\textcolor{blue}{x} - 3)$. Find an expression for the derivative with respect to $\textcolor{blue}{x}$.

[3 marks]

Differentiating this off the bat would be a little tricky (at least, for now), so we’ll have to expand the brackets first.

Expanding the expression gives

$\textcolor{limegreen}{y} = (\textcolor{blue}{x}^\textcolor{red}{2} + 1)(\textcolor{blue}{x} - 3)$

$= \textcolor{blue}{x}^\textcolor{red}{3} - 3\textcolor{blue}{x}^\textcolor{red}{2} + \textcolor{blue}{x} - 3$

So,

$\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \textcolor{red}{3}\textcolor{blue}{x}^{(\textcolor{red}{3} - 1)} - (3 \times \textcolor{red}{2})\textcolor{blue}{x}^{(\textcolor{red}{2} - 1)} + \textcolor{red}{1}\textcolor{blue}{x}^{(\textcolor{red}{1} - 1)} - (3 \times 0)$

$= 3\textcolor{blue}{x}^2 - 6\textcolor{blue}{x} + 1$

A Level   ## Example Questions

For $y = x^4$,

$\dfrac{dy}{dx} = 4x^3$

$f(x) = \dfrac{1}{2x^2} = \dfrac{1}{2}x^{-2}$ gives

$f'(x) = \dfrac{-1}{x^3}$

Simplifying gives

$y = x(x - 3)(x + 5)$

$= x(x^2 + 2x - 15)$

$= x^3 + 2x^2 - 15x$

So

$\dfrac{dy}{dx} = 3x^2 + 4x - 15$

First, expand the brackets:

$y = (x - 6)(x + 2)$

$= x^2 - 4x - 12$

Then, differentiating with respect to $x$ gives

$\dfrac{dy}{dx} = 2x - 4$

When $x = 0$,

$\dfrac{dy}{dx} = (2 \times 0) - 4 = -4$

A Level

A Level

A Level

## You May Also Like... ### A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99 ### A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99 ### Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99