# Differentiation

## Differentiation Revision

**Differentiation**

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

**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}

**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}}}

**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).

**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

## Differentiation Example Questions

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

**[1 mark]**

**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}

**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

So

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

**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