Having discussed in a great amount of detail what a limit
is, we return to our original

question: finding the instantaneous rate of change of a function . We previously
stated that

the average rate of change of a function over an interval Δx is given by

where x_{0} is the base point of our interval, and the second
point of the interval lies at x_{0} + h

(the above formula is called the difference quotient of f at x_{0} with increment
h, because

it is a quotient of a difference). It is worth returning to the graphical
interpretation of the

average rate of change at this point. A line that intersects a curve at two or
more points is

called a secant line. In finding the average rate of change between two points
of a function,

we are finding the slope of the secant line that intersects those two points.

To find the instantaneous rate of change, we need to look at the average rate of
change in

the limit as the length of the interval approaches 0. That is

provided that the limit exists. In this situation, the
notation f'(x_{0}) represents the instantaneous

rate of change of the function f(x) at the point x_{0}. This is a particularly
important

quantity in calculus, so we give it a special name - the derivative of f(x) at
the point x_{0}. The

instantaneous rate of change, or derivative, of f(x) at x_{0} also has an important
graphical

interpretation: f'(x_{0}) is the slope of the line tangent to f(x) at the point
(x_{0}, f(x_{0})). In

this way, we define the line tangent to f(x) at x_{0} as the line that passes
through the point

(x_{0}, f(x_{0})) with slope f'(x_{0}). We define the slope of a function f(x) at a point
x_{0} as the

slope of the tangent line that passes through (x_{0}, f(x_{0})). Now that we have
introduced an

extroardinary amount of notation, let us try to get a hold on it by working
through some

examples.

**Example 1** Let f(x) = x^{2}. Find the equation for the secant line passing through
(2, f(2))

and (2 + h, f(2 + h). Find the equation for the tangent line passing through (2,
f(2)).

** Solution ** In order to find the equation of a line, we need to know either the
slope of the

line and a single point the line passes through, or two points on the line, from
which we can

calculuate the slope. With this information we can use the point slope form (for
a slope m

and point (x_{0}, y_{0}))

Since we know two points through which the secant line
passes, we can find the slope of the

secant line

Now using point-slope form, the above slope, and a single
point on the line (it is customary

to use the base point of the interval - and there is a good reason for doing so
- because we

also want to consider the tangent line at that point) we find

It’s useful to notice what happens as we modify our
interval, by manipulating h. When

we extend our interval to the right, by increasing h, the slope increases and
the y-intercept

decreases. If we let our interval extend to the left of the base point, by
considering negative h

values , the slope decreases and the y- intercept increases . Now let us perform
similar analysis

to find the equation of the tangent line (while reusing as much work above as
possible).

Now we utilize point-slope form once again

These two equations are very similar; in fact, if we
consider the equation for the secant line

in the limit as h → 0, we arrive at the equation of the tangent line.
Graphically, we can see

that as we decrease h, the secant line becomes closer and closer to the tangent
line, and in

the limit as h → 0, the secant line is the tangent line. Thus, we can interpret
the tangent

line in a slightly different way ; that is, the line tangent to f(x) at x_{0} is the
secant line with

base point x_{0} in the limit as the distance between the two points of the
function f(x) the

secant line intersects h → 0.

**Example 2** Let f(x) = mx + b be the equation of an
arbitrary line. Find the equation of

the line tangent to f(x) at an arbitrary point x_{0}.

**Solution** Since the equation of a line passing through a given point is unique,
and f(x)

is the unique line of slope m that crosses through every point it does, we
suspect that the

tangent line should have to be exactly the line f(x) (and in fact, so should any
secant line

passing through the point x_{0}).

so we see that the slope is exactly the same as the line
itself.

Thus, as suspected, the line tangent to a line at any
point is just the line itself.

**Example 3** Find the equation of the line tangent to the function f(x) = x^{3} at x =
0.

**Solution** We begin as usual by looking at the limit as h → 0 of the difference
quotient

Since this line passes through the point (0, 0), the
point-slope equation is very simple

Thus, in this case the tangent line is simply the x -axis.
It is noteworthy that this line actually

intersects the function f(x) = x^{3}, which should dispell the myth that a tangent
line

cannot cross a function.

**Example 4** Find the slope of the line tangent to f(x) = sin(x) at an arbitrary
point x.

**Solution** In finding the derivative of a function at an arbitrary point x, we
define a new

function f'(x), which for any x value gives the corresponding value of the
derivative at that

point. We call this new function f'(x) the derivative of f. Proceeding with the
difference

quotient

We have already seen that the limit

and we can actually use this limit to find the value of
the other one using the half angle

formula cos h = 1 − 2 sin^{2}(h/2).

where in the above analysis we substituted h /2 = θ to
evaluate the limit. Using these two

limits, we find that

This is a very interesting result. When we look at the
rate of change of the function sin(x)

we find that it is the other sinusoidal function cos(x). In fact,

or in other words, the instantaneous rate of change of
sin(x) at any point x is simply the

value of sin(x+ π/2), that is, the value of the same function π/2 to the right of
the point of

interest. Thus, sin(x) is a function which has rate of change directly related
itself.

If we look at the tangent line at x = 2π /3 we find that

and since

we find the equation for this tangent line as

this tangent line will touch the curve at two points,
intersecting the curve at one of them.

If we consider the tangent line at x = π/2 we have

so we have a horizontal line, with equation

y = 1

which touches the curve at an infinite number of points . This should dispell the
myth that

a tangent line can only touch a curve in one place, as we have a tangent line
that touches

the curve in an infinite number of places.

Since we find the derivative at a point using a limit, it
fol lows that if the limit does not exist,

neither will the derivative, nor a nonvertical tangent line (which will be
discussed later). We

say that a function is not differentiable at a point where its derivative does
not exist. A

function is not differentiable at any place it has a:

1. corner. Consider f(x) = |x| which has a corner at x = 0. If we look at the
secant lines

in the limit as h → 0, we see that from the left side the slope of the tangent
line is

approaching −1, and from the right side the slope is approaching 1. Since these
limits

do not agree, the derivative and thus tangent line do not exist at x = 0 (a
similar

analysis will hold for a corner of any function).

2. a cusp. Consider the function at the
point x = 0. From the right side

the secant lines approach a vertical tangent line with slope ∞ and from the left
side

with slope −∞. Since the limit is increasing or decreasing without bound, it
does not

exist, so f(x) is not differentiable at x = 0, the point where it has a cusp.

3. vertical tangent. If the limit of the difference quotient fails to exist
because it approaches

∞ or −∞ from both sides, we say that the function has a vertical tangent at

the point of interest (and since ±∞ are not numbers, the derivative does not
exist).

An example of function with a vertical tangent is
.

4. a discontinuity. If we have a jump discontinuity the secant lines will behave
like when

we have a cusp . A function with a jump discontinuity will have different
behavior on

both sides of the point of interest. For instance, f(x) = |x|/x, f(0) = 1 has a
jump

discontinuity at x = 0. The secant lines approach a horizontal tangent from the
right,

and vertical from the left. Thus, the derivative does not exist at this point
(nor does

a tangent line).