So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in Introduction to Functions and Graphs, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions.
Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course.
First of all, we begin with the assumption that the function , is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginning with the definition of , where is a positive integer—as the product of multiplied by itself times. Later, we defined for a positive integer , and for positive integers and . These definitions leave open the question of the value of where is an arbitrary real number. By assuming the continuity of , we may interpret as where the values of as we take the limit are rational. For example, we may view as the number satisfying
As we see in the following table, .
64 | 77.8802710486 | ||
73.5166947198 | 77.8810268071 | ||
77.7084726013 | 77.9242251944 | ||
77.8162741237 | 78.7932424541 | ||
77.8702309526 | 84.4485062895 | ||
77.8799471543 | 256 |
, the value of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function is differentiable everywhere.
We make one final assumption: that there is a unique value of for which . We define to be this unique value, as we did in Introduction to Functions and Graphs. (Figure) provides graphs of the functions , and . A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of lies somewhere between 2.7 and 2.8. The function is called the natural exponential function. Its inverse, is called the natural logarithmic function.
Figure 1. The graph of is between and .
For a better estimate of , we may construct a table of estimates of for functions of the form . Before doing this, recall that
for values of very close to zero. For our estimates, we choose and to obtain the estimate
.
See the following table.
Estimating a Value of 2 2.7183 2.7 2.719 2.71 2.72 2.718 2.8 2.7182 3
The evidence from the table suggests that .
The graph of together with the line are shown in (Figure). This line is tangent to the graph of at .
Figure 2. The tangent line to at has slope 1.
. Recall that we have assumed that exists. By applying the limit definition to the derivative we conclude that
.
Turning to , we obtain the following.
We see that on the basis of the assumption that is differentiable at is not only differentiable everywhere, but its derivative is
.
For . Thus, we have . (The value of for an arbitrary function of the form , will be derived later.)
Let be the natural exponential function. Then
.
.
Find the derivative of .
Using the derivative formula and the chain rule,
Find the derivative of .
Use the derivative of the natural exponential function, the quotient rule, and the chain rule.
Find the derivative of .
Don’t forget to use the product rule.
A colony of mosquitoes has an initial population of 1000. After days, the population is given by . Show that the ratio of the rate of change of the population, , to the population size, is constant.
First find . By using the chain rule, we have . Thus, the ratio of the rate of change of the population to the population size is given by
.
The ratio of the rate of change of the population to the population size is the constant 0.3.
If describes the mosquito population after days, as in the preceding example, what is the rate of change of after 4 days?
Find .
Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.
and , then
.
, the derivative of is given by
.
and , then . Differentiating both sides of this equation results in the equation
.
Solving for yields
.
Finally, we substitute to obtain
.
We may also derive this result by applying the inverse function theorem, as follows. Since is the inverse of , by applying the inverse function theorem we have
.
Using this result and applying the chain rule to yields
The graph of and its derivative are shown in (Figure).
Figure 3. The function is increasing on . Its derivative is greater than zero on .
Find the derivative of .
Find the derivative of .
At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.
Differentiate: .
Use a property of logarithms to simplify before taking the derivative.
find the derivatives of and for 0, \, b\ne 1" width="90" height="17" /> .
, and let be a differentiable function.
.
,
.
If , then
.
More generally, if , then
.
If , then . It follows that . Thus . Solving for , we have . Differentiating and keeping in mind that is a constant, we see that
.
The derivative in (Figure) now follows from the chain rule.
If , then . Using implicit differentiation, again keeping in mind that is constant, it follows that Solving for and substituting , we see that
.
The more general derivative ((Figure)) follows from the chain rule.
Find the derivative of .
Use the quotient rule and (Figure).
Find the slope of the line tangent to the graph of at .
To find the slope, we must evaluate at . Using (Figure), we see that
.
By evaluating the derivative at , we see that the tangent line has slope
.
Find the slope for the line tangent to at .
Evaluate the derivative at .
and . Unfortunately, we still do not know the derivatives of functions such as or . These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form . It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of . We outline this technique in the following problem-solving strategy.
Find the derivative of .
Use logarithmic differentiation to find this derivative.
Find the derivative of .
This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.
Find the derivative of where is an arbitrary real number.
The process is the same as in (Figure), though with fewer complications.
Use logarithmic differentiation to find the derivative of .
Follow the problem solving strategy.
Find the derivative of .
Use the result from (Figure).
is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
For the following exercises, find for each function.
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For the following exercises, use logarithmic differentiation to find .
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24. [T] Find an equation of the tangent line to the graph of at the point where
. Graph both the function and the tangent line.
25. [T] Find the equation of the line that is normal to the graph of at the point where . Graph both the function and the normal line.
26. [T] Find the equation of the tangent line to the graph of at the point where . (Hint: Use implicit differentiation to find .) Graph both the curve and the tangent line.
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and those where .
on , and on
28. The formula is the formula for a decaying alternating current.
0 | (i) |
(ii) | |
(iii) | |
(iv) | |
(v) | |
(vi) | |
(vii) | |
(viii) | |
(ix) |
29. [T] The population of Toledo, Ohio, in 2000 was approximately 500,000. Assume the population is increasing at a rate of 5% per year.
a. individuals
b. individuals per year
c. 39,737 individuals per year
30. [T] An isotope of the element erbium has a half-life of approximately 12 hours. Initially there are 9 grams of the isotope present.
31. [T] The number of cases of influenza in New York City from the beginning of 1960 to the beginning of 1961 is modeled by the function
,
where gives the number of cases (in thousands) and is measured in years, with corresponding to the beginning of 1960.
a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1963 there were approximately 723 cases of the disease in New York City.
b. At the beginning of 1960 the number of cases of the disease was decreasing at rate of -4.611 thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of -0.2808 thousand per year.
32. [T] The relative rate of change of a differentiable function is given by . One model for population growth is a Gompertz growth function, given by where , and are constants.
For the following exercises, use the population of New York City from 1790 to 1860, given in the following table.
Years since 1790 | Population |
0 | 33,131 |
10 | 60,515 |
20 | 96,373 |
30 | 123,706 |
40 | 202,300 |
50 | 312,710 |
60 | 515,547 |
70 | 813,669 |
33. [T] Using a computer program or a calculator, fit a growth curve to the data of the form .
34. [T] Using the exponential best fit for the data, write a table containing the derivatives evaluated at each year.
35. [T] Using the exponential best fit for the data, write a table containing the second derivatives evaluated at each year.
Years since 1790 | |
---|---|
0 | 69.25 |
10 | 107.5 |
20 | 167.0 |
30 | 259.4 |
40 | 402.8 |
50 | 625.5 |
60 | 971.4 |
70 | 1508.5 |
36. [T] Using the tables of first and second derivatives and the best fit, answer the following questions:
logarithmic differentiation is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly