Summary: Introduction to the derivative

Tools:

Average rate of change over [a, b]: Difference quotient

Given a function $f$, the average rate of change of $f(x)$ over the interval $[a, b]$ is

Average rate of change	$\displaystyle = \frac{\Delta f}{\Delta x} = \frac{f(b) - f(a)}{b - a}$.
	$=$ Slope of line through points $P$ and $Q$ (see the interactive graphic).

We also call this average rate of change the difference quotient of $f$ over the interval $[a, b]$.

Alternative Formulation: Average Rate of Change of over [a, a+h]

(Replace $b$ by $a+h$): The average rate of change of $f(x)$ over the interval $[a,a+h]$ is

Average rate of change $\displaystyle = \frac{f(a+h) - f(a)}{h}$.

Units

The units of the average rate of change are units of $f$ per unit of $x$.

Average rate of change calculators
Use the handy little utilities below to compute the average rate of change of any function you enter.

Examples: Average rate of change over [a, b]: Difference quotient

Let $f(x) = 2x^2-4x+1$. Then the average rate of change of $f$ over the interval $[2,4]$ is

$\displaystyle \frac{f(b) - f(a)}{b - a}$	$\displaystyle = \frac{f(4) - f(2)}{4 - 2}$
	$\displaystyle = \frac{17 - 1}{2} = 8$.

Interpretation: If, say, $x$ represents the year since January 2013, and $f(x)$ represents the accumulated revenue of your company (in millions of dollars) since that time, then the units of measurement of the average rate of change are millions of dollars per year. Thus, your company earned an average annual revenue of &D&8 million per year over the period January 2015 ($t = 2$) to January 2017 ($t = 4$).

Let $f(x) = x^2-x$. Then the average rate of change of $f$ over the interval $[3, 3+h]$ is

$\displaystyle \frac{f(a+h) - f(a)}{h}$	$\displaystyle = \frac{f(3+h) - f(3)}{h}$
	$\displaystyle = \frac{[(3+h)^2 - (3+h)] - [3^2 - 3]}{h}$
	$\displaystyle = \frac{9 + 6h + h^2 - 3 - h - 9 + 3}{h}$
	$\displaystyle = \frac{6h + h^2 - h}{h} = \frac{h(6 + h - 1)}{h}$
	$\displaystyle = 6 + h - 1$

Practice:

Instantaneous Rate of Change at x = a: Derivative

Given a function $f$, we think of the instantaneous rate of change of $f(x)$ at $x = a$ as the average rate of change measured over a very small interval $[a, a+h]$. More precisely, it is defined by taking the limit of the average rates of change over the intervals $[a,a+h]$ as $h$ approaches 0.

Instantaneous rate of change

$\displaystyle = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h}$.

We also call this instantaneous rate of change the derivative of $f$ at $a$, and write it as $f'(a)$ (read "$f$ prime of $a$"). Its units of measurement are units of $f(x)$ per unit of $x$. Thus,

$\displaystyle f'(a) = \lim_{h \to 0}\frac{f(a+h) - f(a)}{h}$.

Notes

$f'(a)$ = Instantaneous rate of change of $f$ at the point $a$. Thus, $f'(x)$ = Instantaneous rate of change of $f$ at the point $x$.
Hence, the derivative $f'(x)$ is a function of $x$, as its value can change with $x$.
As $f'(a)$ is a limit, it may or may not exist. That is, the quantities $\frac{f(a+h) - f(a)}{h}$ may or may not approach a fixed number as $h$ approaches zero. If everything works out fine and the limit exists, then we say that $f$ is differentiable at $a$. Otherwise, we say that $f$ is not differentiable at $a$.
To talk about $f'(a)$, we require that $a$ be an interior point of the domain of $f$; that is, we require that $f(x)$ be defined for all $x$ in some open interval about $x=a$. Thus, for instance, we do not talk about $f'(a)$ if $a$ is an endpoint of the domain of $f$.

Example: Instantaneous Rate of Change at x = a: Derivative

Let $f(x) = 2x^2-4x+1$. You can use the (blue) average rate of change calculator above to verify that, for very small values of $h$, $\dfrac{f(2+h) - f(2)}{h} \approx 4$. Thus, the instantaneous rate of change of $f$ at $x = 2$ is

$f'(2) = 4$. (We will see how to do calculations like this below.)

Interpretation
If, say, $f(x)$ represents the annual profit of your company (in millions of dollars) and $x$ represents the year since January 2015, then the units of measurement of the instantaneous rate of change are millions of dollars per year. Thus, your company's annual profit was increasing at a rate of &D&4 million per year at the start of 2017 ($t = 2$).

Numerical approach

To compute an approximate value of $f'(a)$ (for a given value of $a$) numerically, one can use either:

A table of values
A "quick approximation"

The first approach shows better and better approximations, sometimes allowing you to guess the exact value, while the second method gives a quick estimate.

Using a table

In a table, compute a succession of values of difference quotients

$\dfrac{f(x+h) - f(x)}{h}$ for smaller and smaller values of $h$, and decide what number these values are approaching. (See the example below.)

Quick approximation: Forward difference quotient

Use a single small value for $h$ such as $h = .0001$, and compute the difference quotient:

Forward difference quotient

The smaller $h$, the better the approximation. (See the example below.)

Quick approximation: Balanced difference quotient

The following formula often gives a better estimate of the derivative:

Balanced difference quotient

The smaller $h$, the better the approximation. (See the example below.)

Quick approximation calculators
Enter a function, enter the point $a$, adjust $h$ if you want, and press "Compute".

Examples: Numerical approach

Continuing with the example $f(x) = 2x^2-4x+1$. let us compute an approximate value of $f'(2)$.

Using a table
The difference quotient (with $a = 2$) is

$\dfrac{\color{blue}{f(2+h)} - \color{indianred}{f(2)}}{h} = \dfrac{\color{blue}{2(2+h)^2-4(2+h)+1} - \color{indianred}{2(2)^2-4(2)+1}}{h}$

The following table shows the value of this difference quotient for several values of $h$ approaching zero.

h	1	0.1	0.01	0.001
Difference Quotient	6	4.2	4.02	4.002

As $h$ gets smaller, we see that the value gets closer and closer to 4, so we conclude

$f'(2) \approx 4$.

Using a forward difference quotient

We use the same formula as above, but with $h = 0.0001:$

$\dfrac{\color{blue}{f(2+0.0001)} - \color{indianred}{f(2)}}{0.0001} = \dfrac{\color{blue}{f(2.0001)} - \color{indianred}{f(2)}}{0.0001}$

$\qquad = \dfrac{\color{blue}{2(2.0001)^2-4(2.0001)+1} - \color{indianred}{2(2)^2-4(2)+1}}{h}$

$\qquad = 4.0002.$

(We could have used the little red utility above to do this calculation.)

Notice that the forward difference quotient method does not give the exact answer (4), but the balanced difference quotient will in this case:

Using a balanced difference quotient

This time we use the balanced difference quotient formula with $h = 0.0001:$

$\dfrac{\color{blue}{f(2+0.0001)} - \color{indianred}{f(2-0.0001)}}{0.0002} = \dfrac{\color{blue}{f(2.0001)} - \color{indianred}{f(1.9999)}}{0.0002}$

$\qquad = \dfrac{\color{blue}{2(2.0001)^2-4(2.0001)+1} - \color{indianred}{2(1.9999)^2-4(1.9999)+1}}{h}$

$\qquad = 4.$

(We could have used the little blue utility above to do this calculation.)

Practice:

Velocity

For an object moving in a straight line with position $s(t)$ at time $t$, the average velocity from time $t$ to time $t+h$ is the average rate of change of the position $s(t)$ over $[t, t+h]$ and is therefore given by the difference quotient

$\displaystyle v_{\text{average}} = \frac{s(t+h) - s(t)}{h}$. The instantaneous velocity at time $t$ is given by the instantaneous rate of change of the position $s(t)$ at time $t$ and is therefore given by the derivative $s'(t)$:

$\displaystyle v(t) = s'(t) = \lim_{h \to 0}\frac{s(t+h) - s(t)}{h}$.

Example: Velocity

Practice:

Graphical approach

We saw above that the average rate of change of the function $f$ over the interval $[a, a+h]$ is the slope of the secant line $PQ$ shown in the following graph: As the derivative $f'(a)$ is obtained by taking the limit as $h \to 0$, you can visualize what happens on the graph by pressing the "Make |h| smaller" button: The point $Q$ moves closer and closer to the point $P$, with the result that the secant line begins to resemble the tangent line at the point $P$. Thus, geometrically, the derivative $f'(a)$ is the slope of the tangent to the graph of $f$ at the point where $x = a$.

Examples: Graphical approach

Let $f(x) = 2x^2-4x+1$.

Slope of secant line through points where $x = 2$ and $x = 3$
	= Average rate of change of $f$ over $[2, 3]$ = $6$ (Check on the (blue) average rate of change calculator above.)
Slope of tangent line through the point where $x = 2$
	= Instantaneous rate of change of $f$ at $x = 2$ = $4$ (Check on the (blue) average rate of change calculator above.)

Here is the graph with these two lines shown:

Zooming in:

Algebraic approach

To calculate the derivative of a function algebraically, proceed as follows:

Write down the definition of the derivative,
Substitute for $f(x+h)$ and $f(x)$
Simplify the numerator in order to factor out an $h$. Then cancel the $h$s and take the limit to obtain the answer.

Examples: Algebraic approach

Going back to our first example, $f(x) = 2x^2-4x+1$, Let us now calculate $f'(x)$ algebraically by following the steps above.

$\displaystyle f'(x)$	$\displaystyle {}= \lim_{h \to 0}\frac{\color{red}{f(x+h)} - \color{blue}{f(x)}}{h}$
	$\displaystyle {}= \lim_{h \to 0}\frac{\color{red}{2(x+h)^2-4(x+h)+1} - \color{blue}{(2x^2-4x+1)}}{h}$
	$\displaystyle {}= \lim_{h \to 0}\frac{2x^2+4xh+2h^2-4x-4h+1-2x^2+4x-1}{h}$
	$\displaystyle {}= \lim_{h \to 0}\frac{4xh+2h^2-4h}{h}$
	$\displaystyle {}= \lim_{h \to 0}\frac{h(4x+2h-4)}{h}$
	$\displaystyle {}= \lim_{h \to 0}(4x+2h-4)$
	$\displaystyle {}= 4x-4$

Thus, $f'(x) = 4x-4$.

Practice: