11
Techniques of Differentiation
with Applications
Copyright © Cengage Learning. All rights reserved.
11.5
Derivatives of Logarithmic and Exponential Functions
Copyright © Cengage Learning. All rights reserved.
Derivatives of Logarithmic and Exponential Functions
Derivative of the Natural Logarithm
Recall that ln x = logex.
Quick Example
Derivative of a constant times a function
3
Derivatives of Logarithmic and Exponential Functions
Derivative of the Logarithm with Base b
Notice that, if b = e, we get the same formula
as previously.
Quick Example
4
Derivation of the formulas
and
[ln x] =
[logbx] =
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Derivation of the formulas
To compute
[ln x] =
and
[logbx] =
[ln x], we need to use the definition of the
derivative. We also use properties of the logarithm to help
evaluate the limit.
Definition of the derivative
Algebra
Properties of the logarithm
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Derivation of the formulas
[ln x] =
and
[logbx] =
Algebra
Properties of the logarithm
which we rewrite as
As h → 0+, the quantity x/h is getting large and positive, and
so the quantity in brackets is approaching e, which leaves
us with
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Derivation of the formulas
[ln x] =
and
[logbx] =
The rule for the derivative of logbx follows from the fact that
logbx = ln x/ln b.
If we were to take the derivative of the natural logarithm of
a quantity (a function of x), rather than just x, we would
need to use the chain rule.
8
Derivation of the formulas
[ln x] =
and
[logbx] =
Derivatives of Logarithms of Functions
9
Derivation of the formulas
[ln x] =
and
[logbx] =
Quick Example
u = x2 + 1
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Example 1 – Derivative of Logarithmic Function
Find
Solution:
The calculation thought experiment tells us that we have
the natural logarithm of a quantity, so
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Derivation of the formulas
[ln x] =
and
[logbx] =
The chain rule, gives us the following formulas.
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Derivation of the formulas
[ln x] =
and
[logbx] =
Quick Example
u = x2 + 1
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Derivation of the formulas
[ln x] =
and
[logbx] =
The following shows the correct way of differentiating bx,
beginning with a special case.
Derivative of ex
Quick Example
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Derivation of the formulas
[ln x] =
and
[logbx] =
For bases other than e, we have the following
generalization:
Derivative of bx
If b is any positive number, then
Note that if b = e, we obtain the previous formula.
Quick Example
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Derivation of the formula
[ex] = ex
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Derivation of the formula
[ex] = ex
To find the derivative of ex we use a shortcut.
Write g(x) = ex.
Then ln g(x) = x.
Take the derivative of both sides of this equation to get
or
g(x) = g(x) = ex.
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Derivation of the formula
[ex] = ex
Derivatives of Exponentials of Functions
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Derivation of the formula
[ex] = ex
Quick Example
u = x2 + 1
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Applications
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Example 4 – Epidemics
In the early stages of the AIDS epidemic during the 1980s,
the number of cases in the United States was increasing by
about 50% every 6 months. By the start of 1983, there
were approximately 1,600 AIDS cases in the United States.
Had this trend continued, how many new cases per year
would have been occurring by the start of 1993?
Solution:
To find the answer, we find that t years after the start of
1983 the number of cases is
A = 1,600(2.25t ).
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Example 4 – Solution
cont’d
We are asking for the number of new cases each year. In
other words, we want the rate of change, dA/dt:
= 1,600(2.25)t ln 2.25 cases per year.
At the start of 1993, t = 10, so the number of new cases per
year is
= 1,600(2.25)10 ln 2.25 ≈ 4,300,000 cases per year.
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