Exponential and trigonometric functions play an important role in many models.
Simple models for population growth and the cooling of liquids involve careful
analysis based on rates of change. These rates force the models to use exponential
functions. Slightly more involved models for competing populations and the
motion of a spring give rise to trigonometric functions. We will study this
kind of modelling later in Chapter VI as we investigate models determined
by differential equations.
The Derivatives of Simple Exponential Functions. We
begin with simple exponential functions. At this stage of our studies we
will presume the variables in many models are controlled by an exponential
function like $f(x) = 2^x$. In your prior work in Chapter
I.A and I.B exercises with tangent lines and rates you may have already explored
the derivative of simple exponential functions. We will continue in this section
to describe the derivatives of simple exponential functions more completely.
h
$\frac {2^h-1}h$
1
1
0.01
0.69555501
1e-04
0.6931712
1e-06
0.69314742
1e-08
0.69314718
1e-10
0.69314776
Let's start with $f(x) = 2^x$ and estimate
f '(0). Looking at the graph of this function in Figure *** certainly
supports the belief that f has a tangent line and thus that f
should have a derivative at 0. Consider the data in Table ***. It suggests
that $f'(0)= \lim_{h \to 0} \frac {2^h-1}h \approx 0.69315$
.
x
f(x)=2^{x}
estimate of f '(x)
Estimate of $\frac {f'(x)}{f(x)}$
-4
0.0625
0.043322
0.6931495
-3
0.125
0.086644
0.6931495
-2
0.25
0.173287
0.6931495
-1
0.5
0.346575
0.6931495
0
1
0.69315
0.6931496
1
2
1.386299
0.6931496
2
4
2.772598
0.6931496
3
8
5.545197
0.6931496
4
16
11.09039
0.6931496
Leaving this issue for a moment we turn to the derivative of
$ f$ at $a$, i.e., $f '(a)$. Here it is to
our
advantage to use the characterization of the derivative as $f'(a)=
\lim_{h \to 0} \frac {f(a+h)-f(a)}h $. This leads us to examine the
difference quotient $ \frac {2^{a+h}-2^a}h=2^a\cdot \frac {2^h-1}h$. Now
as we consider the limit as$h \to 0$
we can use the work on f '(0). The fact is that by the definition of the derivative
as $h \to 0$
, $\lim_{h \to 0} \frac {2^h-1}h =f'(0)$ . So, to describe f '(a) in general, we see that $$f'(a)= \lim_{h \to 0}2^a\cdot \frac {2^h-1}h = 2^a\cdot f'(0) \approx 2^a \cdot 0.69315$$
Remarks: 1. This result does seem quite different from any we have
had before because the derivative here is a constant multiple of the value
of the function. In other words the derivative is proportional to the value
of the function. We can check this is several ways to confirm the truth of
the result.
First we can look at the ratio of a numerical estimate of the derivative
at several points and see that even at the estimation level the ratio is
constant. This is done in Figure *** using h =.0001 for the estimation.
Another way to check the result is to draw lines using the function values
times .693 for the slope at various points on the graph to see that these
lines do appear to be tangent. Of course both these techniques do not fully
justify the claim as does the analysis, but they do add to the credibility
in making them sensible based on a variety of experiences.
Finally with technology you might match the graph of a numerical estimation
of the derivative against the graph of .693 times 2^{x} and
notice how close these graphs appear.
2. The general exponential function with a base b ( not equal to 1)
can be treated in the same fashion as the work we have done with the base
2. The result is that when $f(x)=b^x$
then $f '(x)=f(x) f '(0)$. The argument
to justify this is completely analogous to the argument we have just presented
for the base 2.
Try playing with this GeoGebra figure to see the relation between
the graph of the general exponential function and its derivative by moving
the base b on the slider.
3. Here is an argument for why using the base e for an exponential
function should give a function that is its own derivative. [We will treat
this subject again from a modelling point of view in chapter VI.]
From remark #2 we can see that any base b for which $D_x(b^x)=
b^x$ must have its derivative at 0 equal to 1. In terms
of approximations this means that when $h\to 0$, $ \frac {b^h-1}h \approx 1$
, so $b^h-1 \approx h$,
and thus $b^h \approx 1+h$.
Now if we examine the natural logarithm of both sides of this approximation
we have that $\ln(b^h)=h\cdot \ln(b)\approx \ln(h+1)$. Dividing by h and using further the properties of logarithms we
obtain
$$ \ln(b) \approx \frac{\ln(h+1)}h = \ln ( ( 1+h)^{\frac 1h}).$$
Now recall that when $h=\frac 1n \approx 0$
, for large values of $ n$, $$(1+h)^{\frac 1n}=(1+ \frac 1n)^n \approx e$$.
Putting this together with the last statement we have that as $h \to 0$, $\ln(b) \to \ln(e)=1$. But ln (b) is a constant. Therefore, ln (b)
= 1 and it must be that b = e.
Our argument has shown that if there is a base b where the derivative
of $b^x $ at 0 is 1, then that base must be the number $e$.
By reversing the reasoning the converse statement can also be shown true.
That is, we can justify the following core result:
If $f(x)=e^x$ then $f'(0)=1$ and
therefore $f'(x)=e^x=f(x)$
for any real number <$x$.
h
$\frac {e^h-1}h$
1
1.718281828
0.1
1.051709181
0.01
1.005016708
0.001
1.000500167
0.0001
1.000050002
0.00001
1.000005
See Tables *** for the information on $e^x$ comparable
to the work done for $2^x$ and its derivatives.
$x$
$e^x$
$\frac{e^{x+0.0001}-e^x}{0.0001}$
$\frac{\frac{e^{x+0.0001}-e^x}{0.0001}}{e^x}$
-5
0.0067
0.006738
1.000005
-4
0.0183
0.0183157
1.000005
-3
0.0498
0.0497873
1.000005
-2
0.1353
0.135336
1.000005
-1
0.3679
0.3678813
1.000005
0
1
1.000005
1.000005
1
2.7183
2.7182954
1.000005
2
7.3891
7.389093
1.000005
3
20.086
20.085637
1.000005
4
54.598
54.598423
1.000005
5
148.41
148.4139
1.000005
4. What does .69315 have to do with the number 2? This is a
strange coincidence which will make more sense when we delve a little more
deeply into exponential functions in chapter II. If you check on your calculator
you will find that ln(2) is approximately .69314718 which is very close
to the value we found for the derivative of 2^{x} at 0. In
chapter ** we will see that for any base b, D_{x}(b^{x})
= b^{x} ln(b) and then we will be able to complete
the list of derivative for core functions with the derivative of the logarithmic
functions. Here is a summary of this discussion about simple exponential
functions:
Theorem: I.F.6. (The Proportional Derivative of Exponential Functions.)
If $f(x)= b ^x$ where b>0 then $f'(x)= f'(0)*f(x)$ . [We will show later in Chapter II that $f'(0) = ln(b)$.]
In particular if f(x) = e^{x}, then f
'(0) = 1 and therefore f '(x) = e^{x }= f(x)
for any real number x. In operator and Leibniz notations this last
statement is written $D_x(e^x)=e^x$
and $\frac d {dx}(e^x)=e^x$.
Example. I.F.5. Find the derivative of $g(x) = e^{-x}$. Solution: We will look at this from a graphical and an algebraic
viewpoint. From Figure *** as well as the simple algebra we can see that
g(-a)= e^{-(-a)}= e ^{a}.
On the graph this result is seen in the fact that the graph of $ Y = g(x)$
is the mirror image - reflection of the graph of $Y = e^x$. Considering
the tangent line interpretation of the derivative, we see that the tangent
line for the graph of g(x) at the point (-a, g(-a))
will be the mirror image - reflection of the tangent line to the graph of
$e^x$ at the point $(a, e^a)$.
The slope of this line is the derivative of $e^x$ at $a$,
namely $e^a$. The mirror image line will have the opposite
slope, so the derivative of $g$ at $-a$, $g'(-a) =
-e^a$. Now think of any number $x$ as the opposite
of a number, $x=-a$, so $a=-x$. Then the last statement says that $g '(x) = g '(-a) = -e^a =- e^{-x}$.
Here's another way to see this result using the symmetry of the graphs
and algebra: $e^{-x} =\frac 1{e^x} =(\frac 1e)^x$
so g(x) is a simple exponential function with base $b=\frac 1e$
. From the symmetry of the graphs of g(x) and e^{
x}, the slope of the tangent line to the graph of g at (0,1)
is -1, so g '(0)= -1. See Figure *** Thus from the result we have
about derivative of simple exponential functions we have that $g'(x)= g'(0)=-1*e^{-x}$.
The derivative of ln- the natural logarithm.
Here is a brief approach to finding the derivative of the natural
logarithm. Logarithmic functions with other bases may be done similarly.
The logarithmic functions will be treated again in Chapter II as well as
Chapter VI.
So suppose f (x) = ln(x) and we want to find f '(a). We
will use the difference quotient
. The idea is simply to substitute y for f(x) and b
for f(a). The definition of the logarithm gives us that if
then e^{y} = x and similarly if
then e^{ b} = a. It should make sense from your experience
with logarithms that when x approaches a, y approaches
b. You should draw a mapping diagram for ln(x) and e^{x}
to help visualize these last statements.
Now let's consider the difference quotient,
.
It's again time to think about the limit, this time as $y \to b$. The denominator is the difference quotient
for the derivative of e^{x} evaluated at b, so it approaches
e^{b} and so we see that
.
To summarize, we have shown that for $x>0$,
or
.
Note: Notice how this result makes sense with the graphical interpretation.
See Figure ***. When x is close to 0, the slope is quite large, while when
x is large the slope of the tangent line slope seems closer to 0.
Derivatives of Other Logarithms: From your background (or the quick
review box)you may recall that if we know the natural logarithm of a number
x>0 and a base b>0 b¹1,then
the logarithm with the base b is determined by the equation
. Applying the scalar multiple rule to this expression of the general logarithm
function with base b we have that
and
.
Thus for example,
.
Exercises I.F.2. In problems 1 to 3 find the indicated derivatives.
Find f '(1) and f '(-1) using an estimate when appropriate:
f (x)= 2^{ x} + 6x - 3
f (t)= 5^{ }t + t
f (u)= u^{ 20} - 4^{ u}
f (z)= e^{z} + 1/z
f (x)= ln(x)+ 1/x.
Find
when
y = 5 sin (x) - 3 cos (x); y = ln(x^{
2}) + sin (x)
y = cos(x) - 5x - 3 ; y = 3 sin (x)
+ 1/ x
Find f '(x) and g'(x) when
f (x)= e^{x}+e^{-x} ;
g (x)= e^{x}-e^{-x}
f (x)= 2e^{x}+3e^{-x}
; g (x)= 2e^{x}-3e^{-x}
f (x)= Ae^{x}+Be^{-x}
; g (x)= Ae^{x}-Be^{-x}
Notice in Problem 3 how the derivatives of the pairs of functions
f and g are related.
Find a pair of functions f and g where f '(x)
= g(x) and g '(x) = f(x) where
f (0)=1 and g(0) = 0.
Find a pair of functions f and g where f '(x)
= g(x) and g '(x) = f(x) where
f (0)=4 and g(0) = 3.
Find all x where f ' (x) = 0 when
f(x) = sin(x)
f(x) = sin(x) + x
f(x) = cos(x)
f(x) = cos(x) - x
f(x) = sin(x) + cos(x)
Consider F(x) = sin(x) is a probability distribution
for the random variable X on the interval [0, p/2].
Find the probability density at x=p/4,
p/6, and p/3. Explain
briefly why the values for X are more likely to fall closer to p/6 then to p/4 or p/3.
Consider
is a probability distribution for the random variable X on the interval
[0, ln(2)]. Find and estimate as a decimal the probability density at x=.2,
.4, and .6. Explain briefly why the values for X are more likely to fall
closer to .6 then to .4 or .2.
Consider F(x)= ln(x) as a probability distribution
for the random variable X on the interval [1, e]. Find and estimate
as a decimal the probability density at x= 1.5, 2, and 2.5. Explain
briefly why the values for X are more likely to fall closer to 1.5 then
to 2 or 2.5.
Prove Fact I.F.8. ii). Use this to prove Theorem I.F.9.ii).
Find a function P where P'(x) = f(x) for the
following:
Estimate f '(0) for
,
,
and
. Can you see any arithmetical relations between these derivatives and the
derivative of 2^{x} at 0?
Look in some biology or physics book to find models that use a simple
exponential function to describe a variable that can be measured experimentally.
Describe the context for the model and how the model is used to either explain
or predict.
Read some of Napier's
original paper (Text in html ) developing the concept
of a logarithm. Write a short paper discussing his use of a motion interpretation
to explain how the logarithms would help in solving proportions, etc.