### The Derivative of Inverse Functions

The Derivative of Inverse Functions
Inverse Functions Algebraically
In this section we want to consider the derivative of inverse functions. Hopefully
you’re familiar with inverse functions, but in case you’ve forgotten let’s quickly
review the properties of inverse functions that are going to be important to us (if
you would like a more through review see Section 1.5 in the textbook)
First we recall the definition of a one-to-one function:
Definition 1. A function f is one-to-one if f (x1 ) 6= f (x2 ) for all x1 6= x2 in the
domain of f
One-to-one
x1
Many-to-one
f
y1
x1
x2
y2
x2
Domain
Range
Domain
f
y
Range
Now that we have the definition of a one-to-one function in hand, we can define
what an inverse function is:
Definition 2. Suppose f is a one-to-one function with domain D and range R. The
inverse function of f is denoted f −1 and is defined by
f (a) = b ⇔ f −1 (b) = a
The domain of f −1 is R and its range is D.
f −1
f
−1
0
2
7
Domain
1
7
17
−1
0
2
7
Domain
Range
1
3
7
17
Range
1
3
If our function wasn’t one-to-one, then there would exist some b in the range of
f and two inputs a1 6= a2 in the domain of f with f (a1 ) = b = f (a2 ). If this
is the case, what would f −1 (b) be? It could be either a1 or a2 , which would be
ambiguous.
Furthermore, it is important to note that the domain of f −1 is the range of f , and
the range of f −1 is the domain of f .
Great, we have a definition of an inverse function, but how do we find it? Well,
there are two simple steps to finding the derivative of a one-to-one function
y = f (x):
(1) Interchange x and y. For example, if y = 2x + 1 after interchanging
x and y we would have x = 2y + 1.
(2) Solve the resulting equation for y. This is your inverse function!. So,
continuing with our last example, we would get y = 21 x − 12 . This is
the inverse of the function y = 2x + 1.
The Inverse Function Graphically
The graph of f and f −1 are related. Suppose (a, b) is a point on the graph of f .
Then f (a) = b. However, this means that f −1 (b) = a, which implies (b, a) is a
point on the graph of f −1 .
⇔
(a, b)
| {z }
(b, a)
| {z }
point on graph of f −1
point on graph of f
Now, interchanging the x and y coordinate of a point on a graph is the same
as reflecting the point about the line y = x. So, the graph of f −1 is simply the
graph of f reflected about the line y = x. For
example, below if the graph of
√
3
3
−1
f (x) = (x − 1) + 1 and its inverse f (x) = x − 1 + 1.
y = f(x)
y=x
y = f −1 (x)
2
The Derivative of Inverse Functions
Suppose we have a function f (x) and we would like to determine (f −1 )0 (x). Furthermore, suppose we don’t know what f −1 (x) is, so we can’t just take its derivative. What can we do?
Well, if y = f −1 (x), then finding (f −1 )0 (x) is the same as finding dy/dx or y 0 . Now,
since y = f −1 (x), we have
f (y) = x.
This defines y as an implicit function of x, so let’s use implicit differentiation! Doing
so we get
d
d
(f (y)) =
(x) ⇒ f 0 (y)y 0 = 1.
dx
dx
Now, whenever f 0 (y) 6= 0 we have
(f −1 )0 (x) = y 0 =
1
f 0 (y)
=
1
f 0 (f −1 (x))
.
This is a perfectly valid and useful formula, but if we don’t know what f −1 (x) is,
we can’t really use it. To deal with this, we can rewrite the above formula as
1
(f −1 )0 (f (x)) = 0 .
f (x)
Putting this altogether we have the following theorem:
Theorem 3. If f is differentiable and non-zero at x = a, then f −1 is differentiable at
x = f (a) = b and we have
1
(f −1 )0 (b) = 0 −1
f (f (b))
or equivalently
(f −1 )0 (b) =
1
f 0 (a)
Let’s look at a couple examples:
(1) Let f (x) = 2x + 3. Determine (f −1 )0 (−1).
First we note that −1 = 2(−2) + 3 = f (−2). So, by the above theorem
we have
3
1
(f −1 )0 (−1) =
f 0 (−2)
=
1
2
(2) Let f (x) = 2x2 for x ≥ 0. Determine (f −1 )0 (50).
First we note that 50 = 2(5)2 = f (5). So, by the above theorem we
have
1
1
1
(f −1 )0 (50) = 0
=
=
f (5)
4(5)
20
Practice Problems
(1) Let f (x) = x3 − 1. Find the value of (f −1 )0 (x) at x = 7 = f (2).
(2) Let g(t) = 5t2 − 10t + 4 for t ≥ 1. Find the value of (g −1 )0 (t) at t = 4 = g(2).
(3) Suppose the differentiable function y = f (x) has an inverse. Furthermore,
suppose the slope of the line tangent to the graph of y = f (x) at the point
(2, 4) is 1/5. Determine (f −1 )0 (4). (Hint: Think about the relationship between
the graph of a function and its inverse and note that this relationship holds for
tangent lines as well.)
Derivative of Logarithms
One of the most important inverses in all of mathematics is the inverse of the
exponential function f (x) = bx which is defined to be
f −1 (x) = logb (x).
Our goal is to determine
d
(logb (f (x))).
dx
To do this, we recall that by definition every logarithmic equation has an equivalent
exponential equation. Specifically in this case
y = logb (f (x))
|
{z
logrithmic equation
⇔
}
by = f (x)
|
{z
}
exponential equation
4
Note that the exponential equation defines y implicitly as a function of x. So let’s
use implicit differentiation!
d y
d
(b ) =
(f (x))
dx
dx
⇒ by · ln(b) · y 0 = f 0 (x)
f 0 (x)
ln(b)by
⇒ y0 =
⇒ y0 =
f 0 (x)
.
ln(b)f (x)
This gives us the following derivative rule for logarithmic functions:
Proposition 4. Suppose f (x) is a differentiable function and that f (x) > 0. Then
d
f 0 (x)
(logb (f (x))) =
dx
ln(b)f (x)
Let’s look at some examples. Determine the derivatives of the following functions:
(1) f (x) = ln(3/x)
f 0 (x) =
(2) g(s) = log5 (ex )
f 0 (x) =
(3) h(t) = 3 log8 (log2 (t))
0
h (t) =
1
−3/x2
= −
ln(e)(3/x)
x
ex
1
=
x
ln(5)e
ln(5)
1
ln(2)t
ln(8) log2 (t)
=
1
ln(2) ln(8)t log2 (t)
Logrithmic Differentiation
One of the things that is so useful about logarithms is that they transform multiplication and division to addition and subtraction:
logb (f (x)g(x)) = logb (f (x)) + logb (g(x))
5
logb
f (x)
g(x)
!
= logb (f (x)) − logb (g(x))
This could be useful to us since the sum and difference rule are so much easier
than the product and quotient rule.
For example, suppose
x2 + 3
12x
f (x) =
!
x4 − 1
x3
!
It would be a major pain to determine f 0 (x) using the product and quotient rule.
Let’s determine this derivative by first taking the natural log of both sides
ln(f (x)) = ln
x2 + 3
12x
!
x4 − 1
x3
!
!!
x2 + 3
x4 − 1
ln(f (x)) = ln
+ ln
12x
x3
!
ln(f (x)) = ln(x2 + 3) − ln(12x) + ln(x4 − 1) − ln(x3 )
Taking the derivative with respect to x we get:
d
d
(ln(f (x)) =
(ln(x2 + 3) − ln(12x) + ln(x4 − 1) − ln(x3 ))
dx
dx
2x
1
4x3
3
f 0 (x)
= 2
− + 4
−
⇒
f (x)
x +3 x x −1 x
1
4x3
3
2x
− + 4
−
⇒ f 0 (x) = f (x) 2
x +3 x x −1 x
0
⇒ f (x) =
x2 + 3
12x
!
x4 − 1
x3
!
Practice Problems
Determine the derivatives of the following functions:
(2) g(θ) = log3 (1 + θ ln(3))
6
!
1
4x3
3
2x
−
+
−
.
2
4
x +3 x x −1 x
Notice that this is only valid for x > 1 since we need f (x) > 0.
(1) f (x) = ln(x3/2 )
!
(3) h(t) = 3log2 (t)
Use logrithmic differentiation to determine the derivatives of the following functions:
(4) f (t) = t(t + 1)(t + 2)
(5) g(x) =
q
(x2 + 1)(x − 1)2
Derivative of Inverse Trigonometric Functions
Next we want to determine the derivative of the inverse trigonometric functions,
i.e. sin−1 (x), cos−1 (x), tan−1 (x), csc−1 (x), sec−1 (x), cot−1 (x). However, before
we do so we need to make sense of what these functions are.
Defining sin−1 (x)
Consider the graph of f (x) = sin(x)
y = sin(x)
−
π
2
π
2
Immediately from the graph we see that f (x) = sin(x) is not a one-to-one function,
and a function can have a derivative if and only if it is one-to-one, so what do we
do? Well, we restrict the domain of f (x) = sin(x) to make it one-to-one. There
are many ways you can do this, but we’ll choose to restrict the domain to
π
π
− ≤x≤ .
2
2
Notice that since f (x) = sin(x) is π periodic, we don’t lose any information by
making this restriction!
Now that f (x) = sin(x) with a restricted domain is one-to-one, we can define its
inverse f −1 (x) = sin−1 (x):
7
y = sin −1 (x)
π
2
Domain: − 1 ≤ x ≤ 1
Range: −
π
π
≤y≤
2
2
1
−1
−
π
2
Defining cos−1 (x)
Consider the graph of f (x) = cos(x)
y = cos(x)
π
We have to restrict the domain of f (x) = cos(x) to make it one-to-one. We’ll
choose to restrict the domain to
0 ≤ x ≤ π.
Since f (x) = cos(x) is π periodic, we don’t lose any information by making this
restriction!
We can now define f −1 (x) = cos−1 (x):
y = cos −1 (x)
π
Domain: − 1 ≤ x ≤ 1
Range: 0 ≤ y ≤ π
−1
1
8
180
Chapter 3: Differentiation
−1
Defining tan (x)
Below is the graph of f (x) = tan(x)
Inverse Trigonometric Functions
3.9
−
y = tan(x)
π
2
We introduced the six basic inverse trigon
there on the arcsine and arccosine functions
verse trigonometric functions are defined,
tives are computed.
π
2
Inverses of tan x, cot x, sec x, and cs
We’ll restrict the domain of f (x) = tan(x) to
π
π
− <x< .
2
2
−1
−1
We can now define f (x) = tan (x):
Domain: – ∞ ! x ! ∞
Range: – ! ! y ! !
2
2
y
Domain: – ∞ ! x ! ∞
Range:
0! y! !
y
y " cot –1x
y " tan –1x
x
1
2
–1
!
2
–!
2
–2
(a)
Defining csc−1 (x)
Domain:
Range:
!
!
2
–2
The graphs of these four basic inverse trigono
We obtain these graphs by reflecting the gr
(as discussed in Section 1.6) through the lin
gent, arccotangent, arcsecant, and arccoseca
FIGURE 3.40
–1
1
2
x
–2
–1
(b)
Graphs of the arctangent, arccotangent, arcsecant, and arccosecan
Below is the graph of f (x) = csc(x)
The arctangent of x is a radian angle whose
whose cotangent is x, and so forth. The angles
cotangent, secant, and cosecant functions.
DEFINITION
y ! tan"1 x is the number in s -p>2,
y ! cot"1 x is the number in s0, pd fo
9
y ! sec"1 x is the number in [0, p/2) ´
y ! csc"1 x is the number in [-p/2, 0
We use open or half-open intervals when d
tangent, cotangent, secant, and cosecant fun
The graph of y = tan-1 x is symmetric
nctions
y = csc(x)
he six basic inverse trigonometric functions in Section 1.6, but focused
ine and arccosine functions. Here we complete the study of how all six intric functions are defined, graphed, and evaluated, and how their
derivaπ
ed.
n x, cot x, sec x, and csc x
se four basic inverse trigonometric functions are shown again in Figure 3.40.
restrict of
thethe
domain
of f (x)
= csc(x) to functions
graphs by reflectingWe’ll
the graphs
restricted
trigonometric
π
π
Section 1.6) through the line y = x. Let’s take a closer
< xat<the .arctan− look
2
2
nt, arcsecant, and arccosecant functions.
We can now define f −1 (x) = csc−1 (x):
! x!∞
! y! !
Domain: x ! –1 or x " 1
Range: – ! ! y ! ! , y # 0
2
2
y
Domain: x ! –1 or x " 1
Range: 0 ! y ! !, y # !
2
y
!
2
!
y"
cot –1x
!
2
!
2
y " sec
–1x
–2
1
2
x
–2
–1
1
2
y " csc–1x
–1
1
2
x
–!
2
x
(c)
(d)
gent, arcsecant, and arccosecant
functions.
Defining
sec−1 (x)
Below is the graph of f (x) = sec(x)
f x is a radian angle whose tangent is x. The arccotangent
of x is an angle
y = sec(x)
is x, and so forth. The angles belong to the restricted domains of the tangent,
t, and cosecant functions.
N
−
π
2
π
2
x is the number in s -p>2, p>2d for which tan y = x.
x is the number in s0, pd for which cot y = x.
We’ll restrict the domain of f (x) = sec(x) to
x is the number in [0, p/2) ´ (p/2, p] for which sec y = x.
x is the number in [-p/2, 0) ´ (0, p/2] for which csc y = x.
10
half-open intervals when describing the ranges to avoid values where the
nt, secant, and cosecant functions are undefined. (See Figure 3.40.)
f y = tan-1 x is symmetric about the origin because it is a branch of the
that is symmetric about the origin (Figure 3.40a). Algebraically this
tan-1 s -xd = -tan-1 x;
tives are computed.
Inverses of tan x, cot x, sec x, and csc x
The graphs of these four basic inverse trigonometric functions are shown again in Figure 3.40.
We obtain these graphs by reflecting the graphs of the restricted trigonometric functions
(as discussed in Section 1.6) through the line y = x. Let’s take a closer look at the arctan0 < x < π.
gent, arccotangent, arcsecant, and arccosecant functions.
We can now define f −1 (x) = sec−1 (x):
x! ∞
y! !
2
1
Domain: – ∞ ! x ! ∞
Range:
0! y! !
y
!
y " tan
2
cot –1x
!
2
!
2
x
–2
!
2
!
y"
–1x
Domain: x ! –1 or x " 1
Range: – ! ! y ! ! , y # 0
2
2
y
Domain: x ! –1 or x " 1
Range: 0 ! y ! !, y # !
2
y
–1
–2
1
(b)
y " sec
–1x
2
x
–2
–1
1
2
y " csc–1x
–1
1
x
–!
2
x
(c)
2
(d)
−1
raphs of the arctangent,Defining
arccotangent,cot
arcsecant,
(x)and arccosecant functions.
Below is of
thex graph
of f (x)
= cot(x)
The arctangent
angle
whose tangent is x. The arccotangent of x is an angle
y = cot(x)
whose cotangent is x, and so forth. The angles belong to the restricted domains of the tangent,
cotangent, secant, and cosecant functions.
DEFINITION
π
y ! tan"1 x is the number in s -p>2, p>2d for which tan y = x.
y ! cot"1 x is the number in s0, pd for which cot y = x.
We’ll restrict the domain of f (x) = cot(x) to
y ! sec"1 x is the number in [0, p/2) ´ (p/2, p] for which sec y = x.
0 < x < π.
"1
x isdefine
the number
0) ´ (0, p/2] for which csc y = x.
y !
Wecsc
can now
f −1 (x)in=[-p/2,
cot−1 (x):
We use open or half-open intervals when describing the ranges to avoid values where the
tangent, cotangent, secant, and cosecant functions are undefined. (See Figure 3.40.)
The graph of y = tan-1 x is symmetric about the origin because it is a branch of the
graph x = tan y that is symmetric about the origin (Figure 3.40a). Algebraically this
means that
tan-1 s -xd = -tan-1 x;
the arctangent is an odd function. The graph of y = cot-1 x has no such symmetry
(Figure 3.40b). Notice from Figure 3.40a that the graph of the arctangent function has two
11
horizontal asymptotes; one at y = p>2 and the other at y = -p>2.
Inverses of tan x, cot x, sec x, and csc x
The graphs of these four basic inverse trigonometric functions are shown again
We obtain these graphs by reflecting the graphs of the restricted trigonom
(as discussed in Section 1.6) through the line y = x. Let’s take a closer loo
gent, arccotangent, arcsecant, and arccosecant functions.
Domain: – ∞ ! x ! ∞
Range: – ! ! y ! !
2
2
y
!
2
–2
–1
Domain: – ∞ ! x ! ∞
Range:
0! y! !
y
!
y"
cot –1x
!
2
!
2
–1
y " sec –1x
–2
–!
2
(a)
!
2
!
y " tan –1x
x
1
2
–2
Domain: x !
Range: – ! !
2
y
Domain: x ! –1 or x " 1
Range: 0 ! y ! !, y # !
2
y
1
(b)
2
x
–2
–1
1
2
–1
x
(c)
(d
FIGURE
3.40 Graphs
the −1
arctangent,
Derivative
ofofsin
(x) arccotangent, arcsecant, and arccosecant functions.
Let’s determine
The arctangent of x is a radian angle whose tangent is x. The arccotangent o
d
whose cotangent
x, and so forth. The angles belong to the restricted domains
(sin−1is(x)).
dx
cotangent, secant, and cosecant functions.
First note that we must have |x| < 1. Next, if we let y = sin−1 (x) we see that we
want to determine dy/dx or y 0 . We have
DEFINITION
sin(y) = x
"1
y y,
! sotan
is the
number
in s -p>2, p>2d for which tan y = x.
which is an implicit function of
we’llxuse
implicit
differentiation.
number in s0, pd for which cot y = x.
y d!(sin(y))
cot"1 x=is dthe(x)
dx
dx
"1 0
y !
x is=the
⇒sec
cos(y)y
1 number in [0, p/2) ´ (p/2, p] for which sec y = x
y ! csc0 "1 x is1 the number in [-p/2, 0) ´ (0, p/2] for which csc y =
⇒y =
(1)
cos(y)
Since y = sin−1 (x), we can
to get intervals when describing the ranges to avoid va
Wesubstitute
use openback
or half-open
1 secant, and cosecant functions are undefined. (See Figu
tangent,0 cotangent,
.
y =
The graph
of −1
y (x))
= tan-1 x is symmetric about the origin because it is a
cos(sin
graph x = tan y that is symmetric about the origin (Figure 3.40a). Alg
Now, here’s where we’re means
going to
get clever. Remember the identity
that
cos2 (x) + sin2 (x) = 1.
tan-1 s -xd = -tan-1 x;
the arctangent
is get
an odd function. The graph of y = cot-1 x has no s
Well, if we solve this identity
for cos(x) we
(Figure 3.40b). Notice from Figure 3.40a that the graph of the arctangent fu
horizontal asymptotes; one at y = p>2 and the other at y = -p>2.
12
q
cos(x) = ± 1 − sin2 (x).
Now, let’s use this identity in (1). Doing so, we get
1
y0 = ± q
2
1 − sin (sin−1 (x))
⇒ y0 = ± √
1
.
1 − x2
Now, we know that sin−1 (x) is increasing over its domain (see the graph), so we
know it’s derivative is positive. This means we can drop the − and we get
d
1
(sin−1 (x)) = √
dx
1 − x2
If we employ the chain rule, we get the following general proposition
Proposition 5. Let f (x) be a differentiable function whose range is (−1, 1), that is,
|f (x)| < 1 for all x. Then
d
f 0 (x)
.
(sin−1 (f (x)) = q
dx
1 − (f (x))2
Derivative of tan−1 (x)
Let y = tan−1 (x). We want to determine dy/dx or y 0 . We have
tan(y) = x
which is an implicit function of y, so we’ll use implicit differentiation.
d
d
(tan(y)) =
(x)
dx
dx
⇒ sec2 (y)y 0 = 1
1
⇒ y0 =
sec2 (y)
Since y = tan−1 (x), we can substitute back to get
y0 =
1
sec2 (tan−1 (x))
13
(2)
Now, here’s where we’re going to have to be even more clever than last time. Once
again, we’re going to use the identity
cos2 (x) + sin2 (x) = 1.
However, we’re dealing with sec2 (x) here, so we divide both sides of this identity
by cos2 (x) to get
1
cos2 (x) + sin2 (x)
=
2
cos (x)
cos2 (x)
⇒
cos2 (x) sin2 (x)
1
+
=
2
2
cos (x) cos (x)
cos2 (x)
⇒ 1 + tan2 (x) = sec2 (x)
Plugging this into (2), we get
y0 =
1
1 + tan (tan−1 (x))
1
⇒ y0 =
.
1 + x2
2
If we employ the chain rule, we get the following more general proposition:
Proposition 6. Let f (x) be a differentiable function. Then
f 0 (x)
d
(tan−1 (f (x)) =
dx
1 + (f (x))2
Derivative of sec−1 (x)
Let y = sec−1 (x). We want to determine dy/dx or y 0 . First we note that we must
have |x| > 1. Now, since
sec(y) = x
which is an implicit function of y, we’ll use implicit differentiation.
d
d
(sec(y)) =
(x)
dx
dx
⇒ sec(y) tan(y)y 0 = 1
1
⇒ y0 =
sec(y) tan(y)
14
Since sec(x) = y we get
y0 =
1
x tan(y)
(3)
Next, we recall the identity we constructed when we were determining the derivative of tan−1 (x), namely
1 + tan2 (x) = sec2 (x).
Solving for tan(x) we get
q
tan(x) = ± sec2 (x) − 1
Plugging this into (3), we get
1
y0 = ± q
2
x sec (sec−1 (x)) − 1
1
.
⇒ y0 = ± √ 2
x x −1
In this case, we can’t get rid of ±, but we can clean it up a bit. Notice that the
derivative of sec−1 (x) is positive regardless of whether x < −1 or x > 1. Hence,
we can get rid of the ± by
1
d
√
(sec−1 (x)) =
dx
|x| x2 − 1
Once again, if we employ the chain rule, we get the following more general
proposition:
Proposition 7. Let f (x) be a differentiable function with |f (x)| > 1. Then
d
f 0 (x)
q
(sec−1 (f (x)) =
dx
|f (x)| (f (x))2 − 1
Practice Problems
Using arguments similar to those given above, determine the following derivatives:
(1)
d
(cos−1 (x))
dx
(2)
d
(csc−1 (x))
dx
15
(3)
d
(cot−1 (x))
dx