Download 3.5 Derivatives of Inverse Trigonometric Functions

Arkansas Tech University
MATH 2914: Calculus I
Dr. Marcel B. Finan
3.5
Derivatives of Inverse Trigonometric Functions
In this section we introduce the inverse trigonometric functions and then
find their derivatives.
The Inverse Sine Function
The function f (x) = sin x is increasing on the interval [− π2 , π2 ]. Thus, f (x)
is one-to-one and consequently it has an inverse denoted by
f −1 (x) = sin−1 x.
We call this new function the inverse sine function.
From the definition of inverse functions discussed in Section 3.2, we have
the following properties of sin−1 (x) :
(i) Dom(sin−1 x) = Range(sin x) = [−1, 1].
(ii) Range(sin−1 x) = Dom(sin x) = [− π2 , π2 ].
(iii) sin (sin−1 x) = x for all −1 ≤ x ≤ 1.
(iv) sin−1 (sin x) = x for all − π2 ≤ x ≤ π2 .
(v) y = sin−1 x if and only if sin y = x. Using words, the notation y = sin−1 x
gives the angle y whose sine value is x.
Remark 3.5.1
If x is outside the interval [− π2 , π2 ] then we look for the angle y in the interval
[− π2 , π2 ] such that sin x = sin y. In this case, sin−1 (sin x) = y. For example,
−1
sin−1 (sin 5π
(sin π6 ) = π6 .
6 ) = sin
The graph of y = sin−1 x is the reflection of the graph of y = sin x about
the line y = x as shown in Figure 3.5.1.
1
Figure 3.5.1
To find the derivative of the inverse sine, we let y = sin−1 x for −1 ≤ x ≤ 1.
Thus, sin y = x. Applying the implicit differentiation process, we find
1
1
=√
, −1<x<1
cos y
1 − x2
p
√
where cos y = ± 1 − sin2 y = 1 − x2 (draw a right triangle with hypotenuse of length 1, acute angle y, and opposite side of length x). Note we
chose the positive sign since cos y ≥ 0 for − π2 ≤ y ≤ π2 .
Now, if y = sin−1 (u(x)) then by the chain rule, we have
y 0 cos y = 1 =⇒ y 0 =
y0 = √
1
· u0 .
1 − u2
The Inverse Cosine function
In order to define the inverse cosine function, we will restrict the function
f (x) = cos x over the interval [0, π]. There the function is always decreasing.
Therefore f (x) is one-to-one function. Hence, its inverse will be denoted by
f −1 (x) = cos−1 x.
We call cos−1 x the inverse cosine function.
The following are consequences of the definition of inverse functions:
(i) Dom(cos−1 x) = Range(cos x) = [−1, 1].
(ii) Range(cos−1 x) = Dom(cos x) = [0, π].
(iii) cos (cos−1 x) = x for all −1 ≤ x ≤ 1.
(iv) cos−1 (cos x) = x for all 0 ≤ x ≤ π.
(v) y = cos−1 x if and only if cos y = x. Using words, the notation y =
cos−1 x gives the angle y whose cosine value is x.
Remark 3.5.2
If x is outside the interval [0, π] then we look for the angle y in the interval
[0, π] such that cos x = cos y. In this case, cos−1 (cos x) = y. For example,
−1 (cos 5π ) = 5π .
cos−1 (cos 7π
6 ) = cos
6
6
The graph of y = cos−1 x is the reflection of the graph of y = cos x about
the line y = x as shown in Figure 3.5.2.
2
Figure 3.5.2
To find the derivative of the inverse cosine function, we argue in a way
similar to the inverse sine function and find
d
1
(cos−1 x) = − √
, −1<x<1
dx
1 − x2
and
d
1
· u0 .
(cos−1 u(x)) = − √
2
dx
1−u
The Inverse Tangent Function
Although not one-to-one on its full domain, the tangent function is one-toone when restricted to the interval (− π2 , π2 ) since it is increasing there. Thus,
the inverse function exists and is denoted by
f −1 (x) = tan−1 x.
We call this function the inverse tangent function.
As before, we have the following properties:
(i) Dom(tan−1 x) = Range(tan x) = (−∞, ∞).
(ii) Range(tan−1 x) = Dom(tan x) = (− π2 , π2 ).
(iii) tan (tan−1 x) = x for all x.
(iv) tan−1 (tan x) = x for all − π2 < x < π2 .
(v) y = tan−1 x if and only if tan y = x. In words, the notation y = tan−1 x
means that y is the angle whose tangent value is x.
Remark 3.5.3
If x is outside the interval (− π2 , π2 ) and x 6= (2n + 1) π2 , where n is an
integer, then we look for the angle y in the interval (− π2 , π2 ) such that
tan x = tan y. In this case, tan−1 (tan x) = y. For example, tan−1 (tan 5π
6 )=
π
π
−1
tan (tan (− 6 )) = − 6 .
3
The graph of y = tan−1 x is the reflection of y = tan x about the line
y = x as shown in Figure 3.5.3. Note that the lines y = ± π2 are horizontal
asymptotes. That is
lim tan−1 x = − π2
x→−∞
and
lim tan−1 x = π2 .
x→∞
To find the derivative of the inverse tangent, we write y = tan−1 x if and
only if tan y = x. Applying the implicit differentiation process, we find
y 0 sec2 y = 1 =⇒ y 0 = cos2 y =
1
1 + x2
1
where we use the trigonometric identity cos (tan−1 x) = 1+x
2 (right triangle
figure)
Now, for y = tan−1 (u(x)) we apply the chain rule and obtain
d
1
(tan−1 (u(x))) =
· u0 .
dx
1 + u2
Figure 3.5.3
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