# Differentiation from First Principles

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022

## Differentiation from First Principles

The First Principles technique is something of a brute-force method for calculating a derivative – the technique explains how the idea of differentiation first came to being.

A Level

## Finding Derivatives from First Principles

To differentiate from first principles, use the formula

$f'(\textcolor{blue}{x}) = \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{f(\textcolor{blue}{x} + \textcolor{purple}{h}) - f(\textcolor{blue}{x})}{\textcolor{purple}{h}} \right)$

While this might look a little intimidating, it’s pretty easy to understand.

Think about how we describe the gradient between two points for a moment…

$f'(\textcolor{blue}{x}) = \dfrac{d\textcolor{limegreen}{y}}{d\textcolor{blue}{x}} = \dfrac{\text{change in }\textcolor{limegreen}{y}}{\text{change in }\textcolor{blue}{x}}$

Well, we can describe a “change in $\textcolor{limegreen}{y}$” as $f(\textcolor{blue}{x} + \textcolor{purple}{h}) - f(\textcolor{blue}{x})$ and a “change in $\textcolor{blue}{x}$” as the corresponding $\textcolor{blue}{x} + \textcolor{purple}{h} - \textcolor{blue}{x} = \textcolor{purple}{h}$

Surely then, as $\textcolor{purple}{h}$ decreases toward $0$, we find that the value of the gradient tends toward the actual value, $f'(\textcolor{blue}{x})$.

For example, the graph on the right shows the graph $\textcolor{limegreen}{y}=\textcolor{blue}{x}^2$. It also introduces four “chords”, each indicating the gradient between two points on the graph. As the colour transitions from green to purple, the value of $\textcolor{purple}{h}$ is decreasing towards $0$, for the point $(\textcolor{blue}{1},\textcolor{limegreen}{1})$.

$h=2$ gives $f'(x)=4$

$h=1$ gives $f'(x)=3$

$h=0.5$ gives $f'(x)=2.5$

$h=0.1$ gives $f'(x)=2.1$

A Level
A Level

## Example 1: Using the First Principles Technique

Let $f(\textcolor{blue}{x}) = 3\textcolor{blue}{x}^4$. By differentiating from first principles and using the binomial expansion, find $f'(\textcolor{blue}{x})$.

[4 marks]

$f'(x) = \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{f(\textcolor{blue}{x} + \textcolor{purple}{h}) - f(\textcolor{blue}{x})}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{3(\textcolor{blue}{x} + \textcolor{purple}{h})^4 - 3\textcolor{blue}{x}^4}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{3(\textcolor{blue}{x}^{4} + 4\textcolor{blue}{x}^{3}\textcolor{purple}{h} + 6\textcolor{blue}{x}^{2}\textcolor{purple}{h}^{2} + 4\textcolor{blue}{x}\textcolor{purple}{h}^{3} + \textcolor{purple}{h}^{4}) - 3\textcolor{blue}{x}^4}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{3\textcolor{blue}{x}^{4} + 12\textcolor{blue}{x}^{3}\textcolor{purple}{h} + 18\textcolor{blue}{x}^{2}\textcolor{purple}{h}^{2} + 12\textcolor{blue}{x}\textcolor{purple}{h}^{3} + 3\textcolor{purple}{h}^{4} - 3\textcolor{blue}{x}^4}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{12\textcolor{blue}{x}^{3}\textcolor{purple}{h} + 18\textcolor{blue}{x}^{2}\textcolor{purple}{h}^{2} + 12\textcolor{blue}{x}\textcolor{purple}{h}^{3} + 3\textcolor{purple}{h}^{4}}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( 12\textcolor{blue}{x}^{3} + 18\textcolor{blue}{x}^{2}\textcolor{purple}{h} + 12\textcolor{blue}{x}\textcolor{purple}{h}^{2} + 3\textcolor{purple}{h}^{3} \right)$

$= 12\textcolor{blue}{x}^3$

A Level

## Example 2: Using the First Principles Technique (Again)

Let $f(\textcolor{blue}{x}) = (\textcolor{blue}{x} - 1)^2 + 4\textcolor{blue}{x} - 10$. By differentiating from first principles, find $f'(\textcolor{blue}{x})$.

[4 marks]

$f(\textcolor{blue}{x}) = (\textcolor{blue}{x} - 1)^2 + 4\textcolor{blue}{x} - 10 = \textcolor{blue}{x}^2 + 2\textcolor{blue}{x} - 9$

$f'(\textcolor{blue}{x}) = \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{f(\textcolor{blue}{x} + \textcolor{purple}{h}) - f(\textcolor{blue}{x})}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{(\textcolor{blue}{x} + \textcolor{purple}{h})^2 + 2(\textcolor{blue}{x} + \textcolor{purple}{h}) - 9 - \textcolor{blue}{x}^2 - 2\textcolor{blue}{x} + 9}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{\textcolor{blue}{x}^2 + 2\textcolor{purple}{h}\textcolor{blue}{x} + \textcolor{purple}{h}^2 + 2\textcolor{blue}{x} + 2\textcolor{purple}{h} - 9 - \textcolor{blue}{x}^2 - 2\textcolor{blue}{x} + 9}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( \dfrac{2\textcolor{purple}{h}\textcolor{blue}{x} + \textcolor{purple}{h}^2 + 2\textcolor{purple}{h}}{\textcolor{purple}{h}} \right)$

$= \lim\limits_{\textcolor{purple}{h} \to 0} \left( 2\textcolor{blue}{x} + \textcolor{purple}{h} + 2 \right)$

$= 2\textcolor{blue}{x} + 2$

A Level

## Example Questions

$f'(x) = \lim\limits_{h \to 0} \left( \dfrac{f(x + h) - f(x)}{h} \right)$

$= \lim\limits_{h \to 0} \left( \dfrac{x + h - x}{h} \right)$

$= \lim\limits_{h \to 0} \left( \dfrac{h}{h} \right) = \lim\limits_{h \to 0} 1 = 1$, for all $h$ and all $x$.

$\dfrac{dy}{dx} = \lim\limits_{h \to 0} \left( \dfrac{c - c}{h} \right)$

$= \lim\limits_{h \to 0} \left( \dfrac{0}{h} \right)$

$= \lim\limits_{h \to 0} \left( 0 \right)$

$= 0$
$f(x) = (1 + x^2)^2 = 1 + 2x^2 + x^4$

gives

$f'(x) = \lim\limits_{h \to 0} \left( \dfrac{1 + 2(x + h)^2 + (x + h)^4 - 1 - 2x^2 - x^4}{h} \right)$

$= \lim\limits_{h \to 0} \left( \dfrac{1 + 2(x^2 + 2xh + h^2) + (x^4 + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4}) - 1 - 2x^2 - x^4}{h} \right)$

$= \lim\limits_{h \to 0} \left( \dfrac{1 + 2x^2 + 4xh + 2h^2 + x^4 + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4} - 1 - 2x^2 - x^4}{h} \right)$

$= \lim\limits_{h \to 0} \left( \dfrac{4xh + 2h^2 + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4}}{h} \right)$

$= \lim\limits_{h \to 0} \left( 4x + 2h + 4x^{3} + 6x^{2}h + 4xh^{2} + h^{3} \right)$

$= 4x + 4x^{3}$

A Level

A Level

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