Given a function , how do you find the slope of the * tangent line* to the graph at the point ?

(I'm thinking of the * tangent line* as a line
that just skims the graph at , without going through the graph
at that point. This is a vague description, but it will do for now.)

Here's the idea. Pick a point nearby, and draw the
line connecting to . (A line connecting
two points on a graph is called a * secant
line*.)

Thus, h represents how much you "moved over" in the x-direction. The line has slope

If you slide the second point along the graph toward P, the secant line gets closer and closer to the tangent line. Algebraically, this amounts to taking the limit as . Thus, the slope of the tangent at is

* Example.* Let .

(a) Find the slope of the secant line joining to .

(b) Find the slope of the tangent line to at .

In this case, I let in the equation for and compute the limit:

Another form of the tangent line formula is

You can get this formula from the previous one by letting . Then , so gives .

* Example.* Find the slope of the tangent line
to at .

The graph of is a rectangular hyperbola. Notice that by not plugging in a specific number for a, I've obtained a formula which I can use for any a. For example, the slope of the tangent at (i.e. at the point ) is

There is another interpretation of the slopes of the secant line and the tangent line. The slope of the secant line joining to is

This is the *change* in f divided by the *change* in x,
so it represents the * average rate of change* of
f as x goes from a to b (i.e. on the interval ).

What is the slope of the tangent line at a? It represents the * instantaneous rate of change* at .
(Sometimes people get lazy and just say "rate of change" to
mean "*instantaneous* rate of change".)

* Example.* Let

(a) Find the average rate of change of on the interval .

(b) Find the instantaneous rate of change of at .

The instantaneous rate of change of at is at . I'll use the second formula for :

I set and compute the limit:

Thus, the instantaneous rate of change of at is . This means that if f continued to change at the same rate, then for every 4 units that x increased, the function would increase by 1 unit.

Of course, the function does *not* continue to change at the
same rate. In fact, the rate of change of the function changes! ---
the rate of change of the function is a function itself.

Suppose that the function under investigation gives the * position* of an object moving in one dimension.
(Think of something moving left or right along the x-axis, or an
object that is thrown straight upward, and which eventually falls
back to earth.) For instance, suppose that is the position of
the object at time t.

The * average velocity* of the object from
to is the change in position divided by the time
elapsed:

Notice that this is the same as the slope of the secant line to the curve, or the average rate of change.

The * instantaneous velocity* at
is

This is the slope of the tangent line to the curve, or the instantaneous rate of change. You can also use the second formula

Roughly speaking, the instantaneous velocity measures how fast the object is travelling at a particular instant.

* Example.* The position of an object at time t
is

(a) Find the average velocity from to .

(b) Find the average velocity from to .

What does this mean? Notice that and . In other words,
the object moved around from to , but it wound up
back where it started. Since the *net* change in position was
0, the average velocity was 0.

(c) Find the instantaneous velocity when .

I set in

I get

People who have seen calculus before know that is usually called the *
derivative* of at a. It is denoted by

That is, the * derivative* of at is given by

gives the instantaneous rate of change of f at a, or the slope of the tangent line to the graph of at .

The derivative is a function in its own right. Since x is usually used to denote the input variable for a function, it's common to write the definition of the derivative in this form:

f is * differentiable* at x if
exists --- that is, if the limit above is defined.

* Example.* Compute for .

* Example.* Suppose

Is f differentiable at ?

However, the definition of depends on whether h is positive or negative. I need to take the left- and right-hand limits at 1.

The right-hand limit is

The left-hand limit is

Since the left- and right-hand limits agree, the two-sided limit exists. Thus,

This shows that f is differentiable at .

* Example.* A differentiable function is
continuous.

Geometrically, a differentiable function has a tangent line at each point of its graph. You'd suspect that this would rule out gaps, jumps, or vertical asymptotes --- typical discontinuities. In fact, the requirement that a differentiable function have a tangent line at each point means that its graph has no "corners" --- all of the curves and turns are "smooth".

To see algebraically why this result is true, suppose is differentiable at a point c. By definition,

Then

On the one hand, , so the left side is 0. On the other hand, the product of the limits is the limit of the product, so

I can rewrite this as

This says that f is continuous at c.

* Example.* The picture below shows that graph
of a function . Sketch the graph of
.

I'll do each piece separately from left to right. The left hand piece starts out with a small positive slope. The slope increases till it is large and positive at the asymptote.

The piece in the middle starts out with a big positive slope at the left-hand asymptote. It decreases to 0 --- there's a horizontal tangent at the top of the "bump". It continues to decrease, becoming big and negative at the right-hand asymptote.

Finally, the right-hand piece starts out with a big negative slope near the asymptote. As you go out to the right, the slope continues to be negative, but the curve flattens out --- that is, the slope approaches 0.

Putting these observations together produces a picture like this:

* Example.* An often-used rule of thumb is:

The derivative is undefined at a place where a graph has a corner.

Here's an example which illustrates this point. Suppose

Here's the graph:

It looks as though there might be a corner at , but it's hard to tell. Compute the derivative at 0:

since .

Since f is defined in two pieces, I have to compute the limit on the left and right:

The left- and right-hand limits do not agree. Therefore, the two-sided limit is undefined --- f is not differentiable at .

The left- and right-hand limits I computed are sometimes called the
* left- and right-hand derivatives* of f at
. Intuitively, they give the slope of the tangent as you come in from
the left and right, respectively. Thus, the left-hand derivative at 0
is 2 and the right-hand derivative at 0 is 1.

Send comments about this page to: Bruce.Ikenaga@millersville.edu.

Copyright 2005 by Bruce Ikenaga