1
Some Important Functions
This chapter is devoted to derivatives and integrals of exponential, logarithmic, inverse trigonometric and
hyperbolic functions. These functions are useful in a variety of applications.
1.1 Inverse Functions.
Many important functions are obtained from other functions by solving an equation. Often this process
leads to the inverse of the original function.
1.1.1 Concepts.
Every function has one or more inverse functions. If y = f(x) is a function, then solving the equation
y = f(x) for x in terms of y gives one or more inverse functions to y = f(x). Inverse functions are a common
way of creating new functions.
Example 1. Let
F = Fahrenheit temperature
C = Celsius temperature
Then
(1)
C =
5
(F – 32) = f(F)
9
30
This is a typical function. Its graph is at the right. In this example
20
10
F = independent variable
20
40
60
80
100
10
C = dependent variable
A function is a formula that calculates values of the dependent variable from values of the independent
variable. For example, if Fahrenheit temperature is
F = 68
then the function (1) calculates the corresponding Celsius temperature
C =
30
5
(68 – 32) = f(68) = 20
9
20
10
Graphically, we start at F = 68 on the F axis. Then we go vertically to the
5
graph of C = (F – 32) and then we go horizontally to the C axis which is
9
C = 20.
1.1.1 - 1
20
10
40
60
80
100
Suppose we are in a situation where we are given the Celsius temperature C and we want to find the
corresponding Fahrenheit temperature. For example, if the Celsius temperature is
C = 30
then to find the corresponding Fahrenheit temperature we substitute 30 for C in (1) to get
30 =
5
(F – 32)
9
Solving, we get 54 = F – 32 or
F = 86
30
Graphically, we start at C = 30 on the C axis. We go horizontally to the graph
5
of C = (F – 32) and then vertically to the F axis which is F = 86.
9
20
10
20
If we are in a situation where we are repeatedly given the Celsius temperature
40
60
10
and we need to find the corresponding Fahrenheit temperature then our work
will be simpler if we solve the equation (1) for F in terms of C. Doing this we get
(2)
F =
9
C = F – 32 or
5
9
C + 32 = g(C)
5
The result is a new function that gives F in terms of C. We have called it g(C). Now if we are given a
value of C we just plug this into (2) to find the corresponding F. For example, if C = 10 then
F =
9
(10) + 32 = g(10) = 50
5
The function F =
9
5
C + 32 = g(C) is the inverse function to the function C = (F – 32) = f(F). In general,
5
9
Definition 1. Suppose y = f(x) is a function and we solve the equation y = f(x) for x in terms on y to get
x = g(y). Suppose there is at most one solution x for each y. Then x = g(y) is the inverse function to
y = f(x). Sometimes g is denoted by f -1.
Example 1 (continued). If C =
5
9
(F – 32) = f(F) then its inverse function is F = C + 32 = f -1(C).
9
5
Remarks.
1.
The inverse function is the formula for the independent variable in terms of the dependent variable.
2.
The inverse function will be defined for those y for which the equation y = f(x) has a solution.
These are the values of y for which a horizontal line through y on the y axis meets the graph
1.1 - 2
80
100
y = f(x). The set of values for which the equation y = f(x) has a solution x is called the range of the
function f. In Example 1 the inverse function is defined for all C and the range of f is the set of all
real numbers.
3.
The fact that there is just at most one solution x to the equation y = f(x) corresponds to the fact that
a horizontal line through y on the y axis meets the graph in no more than one point. A function
with this property is said to be one-to-one.
4.
The graph of the inverse function is obtained by reflecting the graph of the original function across
the 45 line.
Example 1 (continued). The graphs of C =
5
9
(F – 32) = f(F) and F = C + 32 = f -1(C) are above. As
9
5
you can see the graphs of each one is obtained from the other by reflecting across the 45 line.
If when we solve the equation y = f(x) for x in terms on y there is more than one solution x, then there will
be more than one inverse function. Often there is some physical reason for singling out one of the solutions.
Example 2. Let
x = length of a side of square
A
x
A = area of the square
Then
x
A = x2 = f(x)
(3)
15
In this example x is the independent variable and A is the dependent variable.
If x = 3 then A = 32 = f(3) = 9. See graph at right.
10
5
If we are told the area of a square is
4
2
2
4
A = 16
and we want to find the length of side of a square whose area is 16, then we substitute 16 for A in (1) to get
16 = x2. Solving, we get two solutions x = 4 and x = - 4. This corresponds to the fact that a horizontal line
through A = 16 meets the graph of A = x2 in the two points (x, A) = (4, 16) and (x, A) = (- 4, 16). Each point
5
corresponds to a solution. This is in contrast to Example 1 where the equation C = (F – 32) had only one
9
solution F for a given C. However, since x is the length of the side of a square which is nonnegative we
take the positive solution x = 4.
If we solve the equation (3) for x in terms of A we get
(4)
x =
A = g(A)
and
x = -
A = h(A)
For example, if A = 25 then the mathematical solutions to (3) are
1.1 - 3
x =
25 = g(A) = 5
The functions x =
and
A = g(A) and x = -
x = -
25 = h(A) = - 5
A = h(A) are the inverse functions to the function A = x2 = f(x). In
general,
Definition 2. Suppose y = f(x) is a function and we solve the equation y = f(x) for x in terms on y to get
x = g1(y), x = g2(y), …, x = gn-1(y) and x = gn(y). (So there is as many as n solution x for each y.) Then
x = g1(y), x = g2(y), …, x = gn-1(y) and x = gn(y) are inverse functions to y = f(x).
Example 2. (continued). Returning to example 2, since x is the length of the side of a square which is
nonnegative we choose the first of the two inverse functions to use for computing the length of the side of a
square with a given area, i.e
x =
A = g(A)
Another way to handle the situation where there is more than one solution, is to restrict the domain of the
original function. In Example 2, we can restrict the domain of the function
A = x2 = f(x)
10
to x  0. Now the graph is at the right. Note that it is missing the part for negative x.
So when we solve the equation A = f(x) for x there is just one solution
x =
The function x =
15
5
1
2
A = g(A)
A = g(A) is the inverse function to the function A = x2 = f(x) where x is restricted to be
non-negative. In general,
Definition 3. Suppose y = f(x) is a function and x is restricted to lie in some interval I. (So I is the domain
of the function f.) Suppose f is one-to-one and when we solve the equation y = f(x) for x in terms on y we
get x = g(y) and this is the only solution lying in the interval I. Then x = g(y) is the inverse function to
y = f(x) where x is restricted to lie in I. Sometimes g is denoted by f -1.
Proposition 1. Suppose the function f is one-to-one. Then
(5)
f(f -1(y)) = y
for all y in the range of f
(6)
f -1(f(x)) = x
for all x in the domain of f
(7)
(f -1)-1 = f
Proof. (5) follows form the definition of f -1(y) since x = f -1(y) is the solution of f(x) = y. (6) also follows
from the definition of f -1 since x is the solution of f(x) = f(x). (7) follows from (6) since y = f(x) is the
solution to f -1(y) = x.
Example 3. If f(x) = x2 then (5) says ( A)2 = A and (6) says
1.1 - 4
x2 = x if x  0.
3
4
Example 4. If y = 2x + 3 = f(x) then y – 3 = 2x and x =
y-3
= f -1(y).
2
Here is another example to illustrate the situation where there is more than one inverse function.
Example 5. Suppose y = x2 + 2x = f(x). Find the inverse functions of f along with the restrictions on x that
correspond to their definition and where these inverse functions are defined.
Solutions. A graph (see right) of y = x2 + 2x = f(x) is a parabola opening up. To find
15
the x coordinate of the vertex, we take the derivative and set it equal to zero.
dy
= 2x + 2
dx

dy
= 0
dx
10
5

4
2x + 2 = 0

2
2
x = -1
Putting x = - 1 into y = x2 + 2x = f(x) we get y = - 1 also. So every horizontal line through a value of y larger
than - 1 intersects the graph in two points while every horizontal line through a value of y less than – 1 does
not intersect the graph. So there are two inverse functions. We get one by restricting x so that x  -1. Let's
call this restriction f1(x). We get the other by restricting x so that x  -1. Let's call this restriction f2(x).
Both inverse functions are defined for values of y that are larger than or equal to – 1.
To find a formula for these inverse functions we just solve the equation y = x2 + 2x for x in terms of y. We
bring y on the other side with the x terms to get x2 + 2x – y = 0. Now we solve for x using the quadratic
formula. We get
x =
-2
22 - 4(1)(-y)
- 2  4 + 4y
=
=
2
2
-22 1+y
= -1
2
1+y
So there are two inverse functions, i.e.
2.0
1.5
x = -1-
-1
1 + y = (f1) (y)
and
x = -1+
1.0
-1
1 + y = (f2) (y)
0.5
10
5
5
At the right is the graph of (f2)-1.
10
0.5
1.0
Often we can not find a formula for an inverse function in terms of functions we are
familiar with.
5
Example 6. Let y = x5 + x = f(x). The graph is at the right. The derivative is
dy
= 5x4 + 1.
dx
1.5
1.0
0.5
0.5
5
dy
Since x  0 one has  1. Since the derivative is positive the function is increasing and
dx
4
therefore this function is one-to-one. So this function has a single inverse function. However, in order to
find a formula for it we would have to solve the equation y = x5 + x = f(x) for x in terms of y. This equation
is equivalent to x5 + x – y = 0. It is known that this equation does not have a solution for x in terms of
familiar functions.
1.1 - 5
1.0
1.5
Sometime we define a new function as the inverse function of a familiar function. This is what is done for
1.0
the inverse trigonometric functions
0.5
Example 7. Let y = sin x = f(x). The graph is at the right. The function y = sin x =
f(x) is not one-to-one, but if we restrict x to the interval -

2
x

2
4
2

2
4
0.5
then is one-to-
1.0
one. If – 1  y  1 then a horizontal line through y meets the graph in exactly one
point whose x coordinate satisfies -
2

 x  . It is a solution to y = sin x and it is
2
1.5
1.0
denoted by x = sin-1 y or x = arcsin y. The graph is at the right.
0.5
1.0
0.5
0.5
0.5
-1
Derivatives of Inverse Functions. Suppose y = f(x) is a function and x = f (y) is
dx
an inverse function to y = f(x). We look at two methods to find the derivative =
dy
d -1
f (y)) of x = f -1(y).
dy (
1.0
1.5
Method 1 – Differentiate as usual. If you have an explicit formula for x = f -1(y) in terms of familiar
functions, then you can find its derivative using the usual rules for derivatives.
Example 2 (continued). Suppose y = x2 = f(x) and x =
y = f -1(y). Then
d -1
d
d 1/2
1
1
f (y)) =
( y) = dy
(y ) = 2 y-1/2 =
dy (
dy
2 y
(8)
Example 5 (continued). Suppose y = x2 + 2x = f(x) with the restriction that x  -1 and
x=-1+
1 + y = (f2)-1(y). Then
d
d
1
1
(f )-1(y)) =
(- 1 + (1 + y)1/2) =
(1 + y)-1/2 =
dy ( 2
dy
2
2 1+y
(9)
Method 2 – Implicit differentiation. This works whether or not you have an explicit formula for x = f -1(y)
in terms of familiar functions.
Example 2 (continued). Suppose y = x2 = f(x) and x =
y = f -1(y). Then take the derivative of both sides
2
of y = x with respect to y keeping in mind that x represents a function of y.
d
d 2
dx
dx
1
y =
x  1 = 2x

=

dy ( )
dy ( )
dy
dy
2x
(10)
d -1
1
1
f (y)) =
=
dy (
2x
2 f -1(y)
When you use implicit differentiation to find
d -1
f (y)) you get a formula involving x = f -1(y). If you have
dy (
a formula for x = f -1(y) in terms of familiar functions, then one can substitute for x to get an explicit formula
d
for (f -1(y)) in terms of familiar functions. In Example 9 one has x = y. Putting this in (10) one gets
dy
1.1 - 6
1.0
d -1
1
f (y)) =
dy (
2 y
which is the same as (8).
Example 5 (continued). Suppose y = x2 + 2x = f(x). Then take the derivative of both sides of y = x2 + 2x
with respect to y keeping in mind that x represents a function of y.
d
d 2
dx
dx
dx
y =
(x + 2x)  1 = 2x + 2
= (2 + 2x)
dy ( )
dy
dy
dy
dy
(11)

dx
1
=
dy
2 + 2x
Recall from Example 5, that y = x2 + 2x = f(x) has two inverse functions, i.e.
x = -1-
1 + y = (f1)-1(y)
corresponding to the restriction x  - 1
x = -1+
1 + y = (f2)-1(y)
corresponding to the restriction x  - 1
The formula (11) applies to both inverses. For example, if we place the restriction x  -1 then (11) becomes
(12)
d
1
1
(f )-1(y)) =
=
dy ( 2
2 + 2x
2 + 2 (f2)-1(y)
When you use implicit differentiation to find
d -1
f (y)) you get a formula involving x = f -1(y). If you have
dy (
a formula for x = f -1(y) in terms of familiar functions, then one can substitute for x to get an explicit formula
d
for (f -1(y)) in terms of familiar functions. In this example one has x = 1 + 1 + y. Putting this in (12)
dy
one gets
d
1
(f )-1(y)) =
dy ( 2
2 + 2 (-1 +
1 + y)
=
2
1
1+y
which is the same as (9).
Example 7 (continued). Use implicit differentiation to find the derivative of x = sin-1 y.
Start with the equation y = sin x and take the derivative of both sides with respect to y. Using implicit
differentiation this gives
dy
d
dx
=
[ sin x ] = (cos x)
dy
dy
dy

dx
1
=
=
dy
cos x
1
1 - y2
1
=
1 - sin2 x
1 = (cos x)
1.1 - 7
dx
dy
In this case we were able to eliminate x from the formula for
dx
by using a trig identity. This result is a
dy
formula for the derivative of x = sin-1 y entirely in terms of y. To summarize
d
[ sin-1 y ] =
dy
1
1 - y2
Implicit differentiation can be used to obtain a general formula for
d -1
f (y)).
dy (
Proposition 2. Suppose y = f(x) is a function and x = f -1(y) is an inverse function to y = f(x). Then
(10)
d -1
1
1
f (y)) =
=
dy (
f '(x)
f '(f -1(y))
Proof. Then take the derivative of both sides of y = f(x) with respect to y keeping in mind that x represents
a function of y.
d
d
y =
f(x)
dy ( )
dy ( )
We use the chain rule on the right side. This gives
1 = f '(x)
Solving for
dx
dy
dx
gives
dy
dx
1
=
dy
f '(x)
Since x = f -1(y), we get (10).
Formula (10) is a nice formula. However, in practice is probably just as easy to use the implicit
differentiation method.
1.1 - 8