# Differentiation

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## 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 $\dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}}$, 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

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

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## 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

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

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## 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

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

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

$= 2x + 1$

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).

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## 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,

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

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

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## Differentiation 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 $x = 0$,

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

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