Implicit and Logarithmic Differentiation
Site: | Saylor Academy |
Course: | MA005: Calculus I |
Book: | Implicit and Logarithmic Differentiation |
Printed by: | Guest user |
Date: | Saturday, September 7, 2024, 7:24 PM |
Description
Read this section to learn about implicit and logarithmic differentiation. Work through practice problems 1-6.
Introduction
This short section presents two final differentiation techniques. These two techniques are more specialized than the ones we have already seen and they are used on a smaller class of functions. For some functions, however, one of these may be the only method that works. The idea of each method is straightforward, but actually using each of them requires that you proceed carefully and practice.
Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.10-Implicit-and-Logarithmic-Differentiation.pdf This work is licensed under a Creative Commons Attribution 3.0 License.
Implicit Differentiation
In our work up until now, the functions we needed to differentiate were either given explicitly, such as
, or it was possible to get an explicit formula for them, such as solving
to get
. Sometimes, however, we will have an equation relating
and
which is either difficult or impossible to solve explicitly for
, such as
or
In any case, we can still find
by using implicit differentiation.
The key idea behind implicit differentiation is to assume that is a function of
even if we cannot explicitly solve for
. This assumption does not require any work, but we need to be very careful to treat
as a function when we differentiate and to use the Chain Rule or the Power Rule for Functions.
Example 1: Assume that is a function of
.
Solution: (a) We need the Power Rule for Functions since is a function of
:
(b) We need to use the product rule and the Chain Rule:
(c) We just need to know that and then use the Chain Rule:
Practice 1: Assume that is a function of
. Calculate (a)
and (b)
.
IMPLICIT DIFFERENTIATION:
Example 2: Find the slope of the tangent line to the circle at the point
with and without implicit differentiation.
Solution:
Explicitly: We can solve the equation of the circle for or
.
Since the point is on the top half of the circle (Fig. 1),
and
Implicitly: We differentiate each side of the equation and then solve for
so
.
Solving for , we have
, and, at the point
,
, the same answer we found explicitly.
Practice 2: Find the slope of the tangent line to at the point
with and without implicit differentiation.
In the previous example and practice problem, it was easy to explicitly solve for , and then we could differentiate
to get
. Because we could explicitly solve for
, we had a choice of methods for calculating
. Sometimes, however, we can not explicitly solve for
, and the only way of determining
is implicit differentiation.
Solution: Assuming that is a function of
and differentiating each side of the equation, we get
Practice 3: Determine at
for
.
In practice, the equations may be rather complicated, but if you proceed carefully and step-by-step, implicit differentiation is not difficult. Just remember that must be treated as a function so every time you differentiate a term containing a
you should get something which has a
The algebra needed to solve for
is always easy - if you differentiated correctly the resulting equation will be a linear equation in the variable
.
Example 4: Find the equation of the tangent line to the "tilted' parabola in Fig. 1 at the point
.
Solution: The line goes through the point so we need to find the slope there. Differentiating each side of the equation, we get
Solving for , so the slope at
is
.
Finally, the equation of the line is so
.
Practice 4: Find the points where the graph in Fig. 2 crosses the y-axis, and find the slopes of the tangent lines at those points.
Implicit differentiation is an alternate method for differentiating equations which can be solved explicitly for the function we want, and it is the only method for finding the derivative of a function which we cannot describe explicitly.
Logarithmic Differentiation
In section 2.5 we saw that . If we simply multiply each side by
, we have
. When the logarithm of a function is simpler than the function itself, it is often easier to differentiate the logarithm of
than to differentiate
itself.
Logarithmic Differentiation: .
The derivative of is
times the derivative of the natural logarithm of
. Usually it is easiest to proceed in three steps:
Let's examine what happens when we use this process on an "easy" function, , and a "hard" one,
. Certainly we don't need to use logarithmic differentiation to find the derivative of
, but sometimes it is instructive to try a new algorithm on a familiar function. Logarithmic differentiation is the easiest way to find the derivative of
.
Example 5: Use the pattern to find the derivative of
.
the same result we would obtain using the product rule.
Practice 5: Use logarithmic differentiation to find the derivative of .
We could have differentiated the functions in the example and practice problem without logarithmic differentiation. There are, however, functions for which logarithmic differentiation is the only method we can use. We know how to differentiate to a constant power,
, and a constant to the variable power,
but the function
has both a variable base and a variable power so neither differentiation rule applies to
. We need to use logarithmic differentiation.
Solution: (i)
(ii) .
Then (iii) .
Logarithmic differentiation is an alternate method for differentiating some functions such as products and quotients, and it is the only method we've seen for differentiating some other functions such as variable bases to variable exponents.
Practice Problem Answers
Practice 4: To find where the parabola crosses the y-axis, we can set and solve for the values of
.
Replacing with 0 in
, we have
so
and
The parabola crosses the
-axis approximately at the points
and
.
From Example 4, we know that , so
at the point , the slope is approximately
, and
at the point , the slope is approximately
.