Here are the formulas for the derivatives of and :

I'll derive them at the end. First, I'll give some examples to show how they're used.

* Example.* Compute .

Using the Product Rule, I get

Note: Don't write " " for the second term, since that means " ". Either put the " " on the left side of the log, or write " ".

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

Since , the point of tangency is .

Using the Product Rule, I find that the derivative is

The slope of the tangent at is

The equation of the tangent line is

* Example.* For what values of x does have a horizontal tangent?

Using the Product Rule, I find that

The tangent line is horizontal when the derivative is equal to 0. Set and solve for x:

(To go from the third equation to the fourth, I used the fact that .) There is a horizontal tangent when .

* Example.* Here are several examples using the
Chain Rule. Do you see the pattern?

In all of these cases,

* Example.* (Logarithmic differentiation)
Compute .

You can't use the Power Rule, because the power is x, not a number.

To do this, first, set . Take logs:

Notice that I used the log rule

Now differentiate both sides of . You need the Chain Rule on the left (or the rule from the last example), and the Product Rule on the right:

Note: It's also possible to do this by writing

You can differentiate using the Chain Rule.

* Example.* (Logarithmic differentiation)
Compute .

Now here are some examples using the differentiation rule for .

* Example.* Compute .

Using the Quotient Rule, I get

* Example.* Find the value(s) of x for which
has a horizontal tangent.

Using the Product Rule, I have

Since I'm looking for horizontal tangents, I set and solve for x:

Since can't ever equal 0, I may divide it out. This gives , or .

* Example.* Here are several examples using the
Chain Rule. Do you see the pattern?

In all of these cases,

You can use the formula for the derivative of to derive the formula for differentiating logs to other bases.

Suppose that . The conversion formula for logs says that

Therefore, since is constant,

In other words, the formula for differentiating is

* Example.*

What about exponentials to other bases?

Let . Then

I used the Chain Rule to differentiate .

Notice that when , the formula becomes

This agrees with the differentiation formula for .

* Example.*

Now I'll show where the derivative formulas for and come from. I have to begin with a precise definition.

If , then , the * natural log* of
x, is defined to be the area under the graph of from 1 to x:

Note that since a segment has zero area, .

If , then is the negative of the area under the graph from 1 to x.

This may not be the definition you're familiar with from earlier courses, but it gives the same thing. The advantage is that it's good for doing calculus.

What is ? The definition of the derivative says if , then

Now

So the top of the fraction, , is the shaded area in this picture:

Build two rectangles, one below the curve and one above the curve:

Obviously,

Both rectangles have base .

The height of the big rectangle is (that is, x plugged into the curve equation ).

The height of the little rectangle is .

Therefore,

Plugging these into (*) above and dividing by h, I get

Now take the limit as .

On the left side, .

On the right side, .

By the Squeezing Theorem,

(Actually, I've done this for , so this is the right-hand limit, but a similar argument works for the left-hand limit.)

I've shown that

What about ? In order to derive its derivative, I need a reasonably precise definition of this function.

Look at the picture I used to define :

As x increases, the area increases. This means that is * one-to-one*: If
, then .

This in turn means that has an inverse function. I'll denote the inverse function by .

Since for , I can use the formula for the derivative of an inverse function to get the derivative formula for .

The formula for the derivative of an inverse function says

Since , I have . So

I've shown that

You may still be uncomfortable by the way I defined and , since they may not correspond to the way
they were defined for you in (say) precalculus courses. In order to
justify these definitions better, later on I'll reinterpret the area
definition for as a * definite integral*. Then I'll be able to show that
, by this definition, has all the
properties you'd expect a log function to have, such as

I'll also be able to justify the corresponding properties for , and this will show that these new definitions "agree with" the ones you may have seen earlier.

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

Copyright 2012 by Bruce Ikenaga