The notion of derivative

Introduction to differentiation: Definition of differentiation

The notion of derivative

In the previous section, we defined the notion of difference quotient. We approximated the slope of the tangent line to a graph by taking the difference quotient of the function at the point $a$ with a small difference $h$ . In this section we will see what happens when we make $h$ increasingly smaller. We will discover that the slope of the tangent line to a graph at a point is equal to the limit for $h$ to $0$ of the difference quotient.

But first we will see what happens when $h$ becomes smaller and smaller.

Below you see the graph $f(x)=\frac{1}{4}x^2+2$ and the tangent line $l$ at the point $\rv{2,3.00000}$ .

Approximate the slope of $l$ at the point $\rv{2,3.00000}$ by calculating the difference quotient of $f$ at $2$ with difference $h$ for $h=1$ , $h=\frac{1}{10}$ , $h=\frac{1}{100}$ , $h=\frac{1}{1000}$ , and $h=\frac{1}{10000}$ , respectively. Give your answer to $5$ decimal places.

The difference quotient for $h=1$ is: $1.25000$
The difference quotient for $h=\frac{1}{10}$ is: $1.02500$
The difference quotient for $h=\frac{1}{100}$ is: $1.00250$
The difference quotient for $h=\frac{1}{1000}$ is: $1.00025$
The difference quotient for $h=\frac{1}{10000}$ is: $1.00002$

This follows from the following calculations:

$\begin{array}{r|cl} h&&\text{difference quotient of }f\text{ at }2\text{ with difference }h\\ \hline 1&\phantom{xx}& \frac{f(3)-3.00000}{1}\phantom{0000}=\frac{4.25000-3.00000}{1}=1.25000\\ \frac{1}{10}&&\frac{f(2.10000)-3.00000}{\frac{1}{10}}=\frac{3.10250-3.00000}{\frac{1}{10}}=1.02500\\ \frac{1}{100}&&\frac{f(2.01000)-3.00000}{\frac{1}{100}}=\frac{3.01003-3.00000}{\frac{1}{100}}=1.00250\\ \frac{1}{1000}&&\frac{f(2.00100)-3.00000}{\frac{1}{1000}}=\frac{3.00100-3.00000}{\frac{1}{10}}=1.00025\\ \frac{1}{10000}&&\frac{f(2.00010)-3.00000}{\frac{1}{10000}}=\frac{3.00010-3.00000}{\frac{1}{10000}}=1.00002 \end{array}$

New example

We see that as $h$ becomes smaller and smaller, the difference quotient approximates the slope of the tangent line more and more accurately. This leads to the following definition of the slope of a graph, which we call derivative.

Differentiation

Let $f$ be a function defined on an interval around a point $a$ . If $\lim_{h\to 0}\frac{f(a+h)−f(a)}{h}$ exists, then $f$ is called differentiable at $a$ ; the limit is called the derivative of $f$ at $a$ . This limit is indicated by $f’(a)$ and also by $\dfrac{{\dd f}}{{\dd}x}(a)$ . The act of determining the derivative is called differentiation.

If $f$ is differentiable at all points of an interval $I$ , we say that $f$ differentiable on $I$ . In that case, $f’$ is a function on $I$ .

The value $f'(x)$ is often indicated by $\frac{{\dd }}{{\dd}x}f(x)$ .

If $y$ is a function rule of $f$ , we also write $\left.\frac{{\dd }}{{\dd}x}y\right|_{x=a}$ instead of $f'(a)$ or $\frac{{\dd f }}{{\dd}x}(a)$ .

The number $f'(a)$ is the slope of the graph of $f$ at the point $\rv{a,f(a)}$ .

Often the function $f$ and the function rule $f(x)$ (which is the value of $f$ at any point $x$ ) are used interchangeably. The expressions $f'(x)$ and $\dfrac{\dd}{\dd x}f(x)$ are also used instead of $f'$ .

An example of the use of the vertical bar with $f(x)=x^2+1$ is

$\frac{\dd f}{\dd x}(3)=\left.\frac{\dd}{\dd x}(x^2+1)\right|_{x=3}=\left.(2x)\right|_{x=3}=6\tiny.$

Be aware that in general not every function is differentiable; in this course, this will not be discussed in greater detail.

Using this definition, we can calculate the derivative at a point. This is done in two steps. In the first step, we write down the difference quotient at the point with difference $h$ . In the second step, we let $h$ go to $0$ , that is: we take the limit $h\to 0$ . The examples below show these two steps, first at a specific point, then at a general point $x=a$ .

Calculate the derivative of $f(x)=\frac{1}{4}x^2+1$ at the point $x=2$ .

$1$

First, we calculate the difference quotient of $f$ at $2$ with difference $h$ :

$\begin{array}{rcl}\frac{\Delta y}{\Delta x}&=&\frac{f(2+h)-f(2)}{h}\\ &&\phantom{xx}\color{blue}{\text{definition}}\\ &=&\frac{\frac{1}{4} \cdot (2+h)^2+1-(\frac{1}{4} \cdot (2)^2+1)}{h}\\ &&\phantom{xx}\color{blue}{f(x)=\frac{1}{4}x^2+1}\\ &=&\frac{\frac{1}{4} \cdot (2^2+h^2+2 \cdot 2h)-\frac{1}{4} \cdot (2)^2}{h}\\ &&\phantom{xx}\color{blue}{\text{expanded}}\\&=&\frac{\frac{1}{4} \cdot (h^2+2 \cdot 2h)}{h}={{h}\over{4}}+1\\&&\phantom{xx}\color{blue}{\text{simplified}}\\\end{array}$

Next, we calculate the derivative of $f$ at the point $x=2$ as the limit for $h \to 0$ of the difference quotient:

$f'(2)= \lim_{h \to 0}\left ({{h}\over{4}}+1\right)=1\tiny.$

New example