Differentiation

A LevelAQAEdexcelOCR

Differentiation

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

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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 $dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}}$ , is the result of differentiating $textcolor{limegreen}{y}$ with respect to $textcolor{blue}{x}$ , given by the equation

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

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

$dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = textcolor{red}{2}textcolor{blue}{x}^{textcolor{red}{2} – 1} = textcolor{red}{2}textcolor{blue}{x}$

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

$dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = textcolor{red}{-1}textcolor{blue}{x}^{textcolor{red}{-1} – 1} = – textcolor{blue}{x}^{-2} = dfrac{-1}{textcolor{blue}{x}^2}$

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

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

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

$dfrac{dtextcolor{limegreen}{y}}{dtextcolor{blue}{x}} = textcolor{red}{2}textcolor{blue}{x}^{textcolor{red}{2} – 1} + textcolor{red}{1}textcolor{blue}{x}^{textcolor{red}{1} – 1} – (1 times textcolor{red}{0})textcolor{blue}{x}^{textcolor{red}{0} – 1}$

$= 2textcolor{blue}{x} + 1 – 0$

$= 2x + 1$

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

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

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

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

So,

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

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

Differentiation Example Questions

Question 1: Write the derivative of $y = x^4$ with respect to $x$ .

[1 mark]

For $y = x^4$ ,

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

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Question 2: Given that $f(x) = dfrac{1}{2x^2}$ , find the derivative with respect to $x$ .

[2 marks]

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

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

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Question 3: By first simplifying the expression $y = x(x – 3)(x + 5)$ , find the derivative with respect to $x$ .

[3 marks]

Simplifying gives

$y = x(x – 3)(x + 5)$

$= x(x^2 + 2x – 15)$

$= x^3 + 2x^2 – 15x$

$dfrac{dy}{dx} = 3x^2 + 4x – 15$

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Question 4: For the graph of $y = (x – 6)(x + 2)$ , find the gradient when $x = 0$ .

[3 marks]

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 $x = 0$ ,

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

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Additional Resources

Exam Tips Cheat Sheet

A Level

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Formula Booklet

A Level

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Specification Points Covered

G1 – Understand and use the derivative of $f(x)$ as the gradient of the tangent to the graph of $y=f(x)$ at a general point $(x,y)$ ; the gradient of the tangent as a limit; interpretation as a rate of change; sketching the gradient function for a given curve; second derivatives; differentiation from first principles for small positive integer powers of $x$
G2 – Differentiate $x^{n}$ , for rational values of $n$ , and related constant multiples, sums and differences