Section 3.5
Inverse Trigonometric
Functions
V63.0121.041, Calculus I
New York University
November 1, 2010
Announcements
Mi...
. . . . . .
Announcements
Midterm grades have been
submitted
Quiz 3 this week in
recitation on Section 2.6,
2.8, 3.1, 3.2
...
. . . . . .
Objectives
Know the definitions,
domains, ranges, and
other properties of the
inverse trignometric
functions: ...
. . . . . .
What is an inverse function?
Definition
Let f be a function with domain D and range E. The inverse of f is the...
. . . . . .
What is an inverse function?
Definition
Let f be a function with domain D and range E. The inverse of f is the...
. . . . . .
What functions are invertible?
In order for f−1
to be a function, there must be only one a in D
corresponding ...
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arcta...
. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin
.
.−
π
2
.
.
...
. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin
.
.
.
.−
π
2
...
. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin
.
.
.
.−
π
2
...
. . . . . .
arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
. .x
.y
.sin
.
.
.
.−
π
2
...
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos
.
.0
.
.π
V63.012...
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos
.
.
.
.0
.
.π
V63...
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos
.
.
.
.0
.
.π
.y ...
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos
.
.
.
.0
.
.π
.
....
. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
. .x
.y
.tan
.−
3π
2
.−...
. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
. .x
.y
.tan
.−
3π
2
.−...
. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
. .x
.y
.tan
.−
3π
2
.−...
. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
. .x
.y
.arctan
.−
π
2
...
. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to
[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−
3π
2
.−
π...
. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to
[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−
3π
2
.−
π...
. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to
[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−
3π
2
.−
π...
. . . . . .
arcsec
Arcsecant is the inverse of secant after restriction to
[0, π/2) ∪ (π, 3π/2].
. .x
.y
.
.
.
.
.
π
2
.
3...
. . . . . .
Values of Trigonometric Functions
x 0
π
6
π
4
π
3
π
2
sin x 0
1
2
√
2
2
√
3
2
1
cos x 1
√
3
2
√
2
2
1
2
0
tan ...
. . . . . .
Check: Values of inverse trigonometric functions
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)
V63.01...
. . . . . .
Check: Values of inverse trigonometric functions
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)
Soluti...
. . . . . .
What is arctan(−1)?
. .
.
.
.3π/4
.
.−π/4
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Fun...
. . . . . .
What is arctan(−1)?
. .
.
.
.3π/4
.
.−π/4
.sin(3π/4) =
√
2
2
.cos(3π/4) = −
√
2
2
Yes, tan
(
3π
4
)
= −1
V63.0...
. . . . . .
What is arctan(−1)?
. .
.
.
.3π/4
.
.−π/4
.sin(3π/4) =
√
2
2
.cos(3π/4) = −
√
2
2
Yes, tan
(
3π
4
)
= −1
But, ...
. . . . . .
What is arctan(−1)?
. .
.
.
.3π/4
.
.−π/4
.sin(π/4) = −
√
2
2
.cos(π/4) =
√
2
2
Yes, tan
(
3π
4
)
= −1
But, th...
. . . . . .
What is arctan(−1)?
. .
.
.
.3π/4
.
.−π/4
.sin(π/4) = −
√
2
2
.cos(π/4) =
√
2
2
Yes, tan
(
3π
4
)
= −1
But, th...
. . . . . .
Check: Values of inverse trigonometric functions
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)
Soluti...
. . . . . .
Check: Values of inverse trigonometric functions
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)
Soluti...
. . . . . .
Caution: Notational ambiguity
..sin2
x = (sin x)2
.sin−1
x = (sin x)−1
sinn
x means the nth power of sin x, ex...
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arcta...
. . . . . .
The Inverse Function Theorem
Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f′
(a) ̸...
. . . . . .
The Inverse Function Theorem
Theorem (The Inverse Function Theorem)
Let f be differentiable at a, and f′
(a) ̸...
. . . . . .
Illustrating the Inverse Function Theorem
.
.
Example
Use the inverse function theorem to find the derivative ...
. . . . . .
Illustrating the Inverse Function Theorem
.
.
Example
Use the inverse function theorem to find the derivative ...
. . . . . .
Illustrating the Inverse Function Theorem
.
.
Example
Use the inverse function theorem to find the derivative ...
. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=...
. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=...
. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=...
. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=...
. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=...
. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=...
. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=...
. . . . . .
Graphing arcsin and its derivative
The domain of f is [−1, 1],
but the domain of f′
is
(−1, 1)
lim
x→1−
f′
(x)...
. . . . . .
Composing with arcsin
Example
Let f(x) = arcsin(x3
+ 1). Find f′
(x).
V63.0121.041, Calculus I (NYU) Section 3...
. . . . . .
Composing with arcsin
Example
Let f(x) = arcsin(x3
+ 1). Find f′
(x).
Solution
We have
d
dx
arcsin(x3
+ 1) =
1...
. . . . . .
Derivation: The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin y
dy
dx
= 1 =⇒
dy
dx
=
1
− sin...
. . . . . .
Derivation: The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin y
dy
dx
= 1 =⇒
dy
dx
=
1
− sin...
. . . . . .
Graphing arcsin and arccos
..|
.−1
.|
.1
.
..arcsin
.
..arccos
V63.0121.041, Calculus I (NYU) Section 3.5 Inve...
. . . . . .
Graphing arcsin and arccos
..|
.−1
.|
.1
.
..arcsin
.
..arccos
Note
cos θ = sin
(π
2
− θ
)
=⇒ arccos x =
π
2
−...
. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y...
. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y...
. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y...
. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y...
. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y...
. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y...
. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y...
. . . . . .
Graphing arctan and its derivative
. .x
.y
.arctan
.
1
1 + x2
.π/2
.−π/2
The domain of f and f′
are both (−∞, ...
. . . . . .
Composing with arctan
Example
Let f(x) = arctan
√
x. Find f′
(x).
V63.0121.041, Calculus I (NYU) Section 3.5 I...
. . . . . .
Composing with arctan
Example
Let f(x) = arctan
√
x. Find f′
(x).
Solution
d
dx
arctan
√
x =
1
1 +
(√
x
)2
d
d...
. . . . . .
Derivation: The derivative of arcsec
Try this first.
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigon...
. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= ...
. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= ...
. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= ...
. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= ...
. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= ...
. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= ...
. . . . . .
Derivation: The derivative of arcsec
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= ...
. . . . . .
Another Example
Example
Let f(x) = earcsec 3x
. Find f′
(x).
V63.0121.041, Calculus I (NYU) Section 3.5 Invers...
. . . . . .
Another Example
Example
Let f(x) = earcsec 3x
. Find f′
(x).
Solution
f′
(x) = earcsec 3x
·
1
3x
√
(3x)2 − 1
·...
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arcta...
. . . . . .
Application
Example
One of the guiding principles of
most sports is to “keep your
eye on the ball.” In basebal...
. . . . . .
Application
Example
One of the guiding principles of
most sports is to “keep your
eye on the ball.” In basebal...
. . . . . .
Solution
Let y(t) be the distance from the ball to home plate, and θ the angle the
batter’s eyes make with hom...
. . . . . .
Solution
Let y(t) be the distance from the ball to home plate, and θ the angle the
batter’s eyes make with hom...
. . . . . .
Solution
Let y(t) be the distance from the ball to home plate, and θ the angle the
batter’s eyes make with hom...
. . . . . .
Solution
Let y(t) be the distance from the ball to home plate, and θ the angle the
batter’s eyes make with hom...
. . . . . .
Summary
y y′
arcsin x
1
√
1 − x2
arccos x −
1
√
1 − x2
arctan x
1
1 + x2
arccot x −
1
1 + x2
arcsec x
1
x
√
x2...
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Lesson16 -inverse_trigonometric_functions_041_slides

  1. 1. Section 3.5 Inverse Trigonometric Functions V63.0121.041, Calculus I New York University November 1, 2010 Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations . . . . . .
  2. 2. . . . . . . Announcements Midterm grades have been submitted Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2 Thank you for the evaluations V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 2 / 32
  3. 3. . . . . . . Objectives Know the definitions, domains, ranges, and other properties of the inverse trignometric functions: arcsin, arccos, arctan, arcsec, arccsc, arccot. Know the derivatives of the inverse trignometric functions. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 3 / 32
  4. 4. . . . . . . What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function f−1 defined by: f−1 (b) = a, where a is chosen so that f(a) = b. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
  5. 5. . . . . . . What is an inverse function? Definition Let f be a function with domain D and range E. The inverse of f is the function f−1 defined by: f−1 (b) = a, where a is chosen so that f(a) = b. So f−1 (f(x)) = x, f(f−1 (x)) = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 4 / 32
  6. 6. . . . . . . What functions are invertible? In order for f−1 to be a function, there must be only one a in D corresponding to each b in E. Such a function is called one-to-one The graph of such a function passes the horizontal line test: any horizontal line intersects the graph in exactly one point if at all. If f is continuous, then f−1 is continuous. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 5 / 32
  7. 7. . . . . . . Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 6 / 32
  8. 8. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . .− π 2 . . π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
  9. 9. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . . . .− π 2 . . π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
  10. 10. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . . . .− π 2 . . π 2 .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
  11. 11. . . . . . . arcsin Arcsin is the inverse of the sine function after restriction to [−π/2, π/2]. . .x .y .sin . . . .− π 2 . . π 2 . ..arcsin The domain of arcsin is [−1, 1] The range of arcsin is [ − π 2 , π 2 ] V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 7 / 32
  12. 12. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . .0 . .π V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
  13. 13. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . . . .0 . .π V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
  14. 14. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . . . .0 . .π .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
  15. 15. . . . . . . arccos Arccos is the inverse of the cosine function after restriction to [0, π] . .x .y .cos . . . .0 . .π . ..arccos The domain of arccos is [−1, 1] The range of arccos is [0, π] V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 8 / 32
  16. 16. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .tan .− 3π 2 .− π 2 . π 2 . 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
  17. 17. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .tan .− 3π 2 .− π 2 . π 2 . 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
  18. 18. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .tan .− 3π 2 .− π 2 . π 2 . 3π 2 .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
  19. 19. . . . . . . arctan Arctan is the inverse of the tangent function after restriction to [−π/2, π/2]. . .x .y .arctan .− π 2 . π 2 The domain of arctan is (−∞, ∞) The range of arctan is ( − π 2 , π 2 ) lim x→∞ arctan x = π 2 , lim x→−∞ arctan x = − π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 9 / 32
  20. 20. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y .sec .− 3π 2 .− π 2 . π 2 . 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
  21. 21. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y .sec .− 3π 2 .− π 2 . π 2 . 3π 2 . . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
  22. 22. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y .sec .− 3π 2 .− π 2 . π 2 . 3π 2 . . .y = x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
  23. 23. . . . . . . arcsec Arcsecant is the inverse of secant after restriction to [0, π/2) ∪ (π, 3π/2]. . .x .y . . . . . π 2 . 3π 2 The domain of arcsec is (−∞, −1] ∪ [1, ∞) The range of arcsec is [ 0, π 2 ) ∪ (π 2 , π ] lim x→∞ arcsec x = π 2 , lim x→−∞ arcsec x = 3π 2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 10 / 32
  24. 24. . . . . . . Values of Trigonometric Functions x 0 π 6 π 4 π 3 π 2 sin x 0 1 2 √ 2 2 √ 3 2 1 cos x 1 √ 3 2 √ 2 2 1 2 0 tan x 0 1 √ 3 1 √ 3 undef cot x undef √ 3 1 1 √ 3 0 sec x 1 2 √ 3 2 √ 2 2 undef csc x undef 2 2 √ 2 2 √ 3 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 11 / 32
  25. 25. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
  26. 26. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) Solution π 6 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 12 / 32
  27. 27. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
  28. 28. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(3π/4) = √ 2 2 .cos(3π/4) = − √ 2 2 Yes, tan ( 3π 4 ) = −1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
  29. 29. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(3π/4) = √ 2 2 .cos(3π/4) = − √ 2 2 Yes, tan ( 3π 4 ) = −1 But, the range of arctan is( − π 2 , π 2 ) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
  30. 30. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(π/4) = − √ 2 2 .cos(π/4) = √ 2 2 Yes, tan ( 3π 4 ) = −1 But, the range of arctan is( − π 2 , π 2 ) Another angle whose tangent is −1 is − π 4 , and this is in the right range. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
  31. 31. . . . . . . What is arctan(−1)? . . . . .3π/4 . .−π/4 .sin(π/4) = − √ 2 2 .cos(π/4) = √ 2 2 Yes, tan ( 3π 4 ) = −1 But, the range of arctan is( − π 2 , π 2 ) Another angle whose tangent is −1 is − π 4 , and this is in the right range. So arctan(−1) = − π 4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 13 / 32
  32. 32. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) Solution π 6 − π 4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
  33. 33. . . . . . . Check: Values of inverse trigonometric functions Example Find arcsin(1/2) arctan(−1) arccos ( − √ 2 2 ) Solution π 6 − π 4 3π 4 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 14 / 32
  34. 34. . . . . . . Caution: Notational ambiguity ..sin2 x = (sin x)2 .sin−1 x = (sin x)−1 sinn x means the nth power of sin x, except when n = −1! The book uses sin−1 x for the inverse of sin x, and never for (sin x)−1 . I use csc x for 1 sin x and arcsin x for the inverse of sin x. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 15 / 32
  35. 35. . . . . . . Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 16 / 32
  36. 36. . . . . . . The Inverse Function Theorem Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and (f−1 )′ (b) = 1 f′ (f−1 (b)) In Leibniz notation we have dx dy = 1 dy/dx V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
  37. 37. . . . . . . The Inverse Function Theorem Theorem (The Inverse Function Theorem) Let f be differentiable at a, and f′ (a) ̸= 0. Then f−1 is defined in an open interval containing b = f(a), and (f−1 )′ (b) = 1 f′ (f−1 (b)) In Leibniz notation we have dx dy = 1 dy/dx Upshot: Many times the derivative of f−1 (x) can be found by implicit differentiation and the derivative of f: V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 17 / 32
  38. 38. . . . . . . Illustrating the Inverse Function Theorem . . Example Use the inverse function theorem to find the derivative of the square root function. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
  39. 39. . . . . . . Illustrating the Inverse Function Theorem . . Example Use the inverse function theorem to find the derivative of the square root function. Solution (Newtonian notation) Let f(x) = x2 so that f−1 (y) = √ y. Then f′ (u) = 2u so for any b > 0 we have (f−1 )′ (b) = 1 2 √ b V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
  40. 40. . . . . . . Illustrating the Inverse Function Theorem . . Example Use the inverse function theorem to find the derivative of the square root function. Solution (Newtonian notation) Let f(x) = x2 so that f−1 (y) = √ y. Then f′ (u) = 2u so for any b > 0 we have (f−1 )′ (b) = 1 2 √ b Solution (Leibniz notation) If the original function is y = x2 , then the inverse function is defined by x = y2 . Differentiate implicitly: 1 = 2y dy dx =⇒ dy dx = 1 2 √ x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 18 / 32
  41. 41. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
  42. 42. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
  43. 43. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . .1 .x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
  44. 44. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . .1 .x . .y = arcsin x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
  45. 45. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: . .1 .x . .y = arcsin x . √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
  46. 46. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: cos(arcsin x) = √ 1 − x2 . .1 .x . .y = arcsin x . √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
  47. 47. . . . . . . Derivation: The derivative of arcsin Let y = arcsin x, so x = sin y. Then cos y dy dx = 1 =⇒ dy dx = 1 cos y = 1 cos(arcsin x) To simplify, look at a right triangle: cos(arcsin x) = √ 1 − x2 So d dx arcsin(x) = 1 √ 1 − x2 . .1 .x . .y = arcsin x . √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 19 / 32
  48. 48. . . . . . . Graphing arcsin and its derivative The domain of f is [−1, 1], but the domain of f′ is (−1, 1) lim x→1− f′ (x) = +∞ lim x→−1+ f′ (x) = +∞ ..| .−1 .| .1 . ..arcsin . 1 √ 1 − x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 20 / 32
  49. 49. . . . . . . Composing with arcsin Example Let f(x) = arcsin(x3 + 1). Find f′ (x). V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
  50. 50. . . . . . . Composing with arcsin Example Let f(x) = arcsin(x3 + 1). Find f′ (x). Solution We have d dx arcsin(x3 + 1) = 1 √ 1 − (x3 + 1)2 d dx (x3 + 1) = 3x2 √ −x6 − 2x3 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 21 / 32
  51. 51. . . . . . . Derivation: The derivative of arccos Let y = arccos x, so x = cos y. Then − sin y dy dx = 1 =⇒ dy dx = 1 − sin y = 1 − sin(arccos x) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
  52. 52. . . . . . . Derivation: The derivative of arccos Let y = arccos x, so x = cos y. Then − sin y dy dx = 1 =⇒ dy dx = 1 − sin y = 1 − sin(arccos x) To simplify, look at a right triangle: sin(arccos x) = √ 1 − x2 So d dx arccos(x) = − 1 √ 1 − x2 . .1 . √ 1 − x2 .x . .y = arccos x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 22 / 32
  53. 53. . . . . . . Graphing arcsin and arccos ..| .−1 .| .1 . ..arcsin . ..arccos V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
  54. 54. . . . . . . Graphing arcsin and arccos ..| .−1 .| .1 . ..arcsin . ..arccos Note cos θ = sin (π 2 − θ ) =⇒ arccos x = π 2 − arcsin x So it’s not a surprise that their derivatives are opposites. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 23 / 32
  55. 55. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
  56. 56. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
  57. 57. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . .x .1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
  58. 58. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . .x .1 . .y = arctan x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
  59. 59. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: . .x .1 . .y = arctan x . √ 1 + x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
  60. 60. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: cos(arctan x) = 1 √ 1 + x2 . .x .1 . .y = arctan x . √ 1 + x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
  61. 61. . . . . . . Derivation: The derivative of arctan Let y = arctan x, so x = tan y. Then sec2 y dy dx = 1 =⇒ dy dx = 1 sec2 y = cos2 (arctan x) To simplify, look at a right triangle: cos(arctan x) = 1 √ 1 + x2 So d dx arctan(x) = 1 1 + x2 . .x .1 . .y = arctan x . √ 1 + x2 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 24 / 32
  62. 62. . . . . . . Graphing arctan and its derivative . .x .y .arctan . 1 1 + x2 .π/2 .−π/2 The domain of f and f′ are both (−∞, ∞) Because of the horizontal asymptotes, lim x→±∞ f′ (x) = 0 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 25 / 32
  63. 63. . . . . . . Composing with arctan Example Let f(x) = arctan √ x. Find f′ (x). V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
  64. 64. . . . . . . Composing with arctan Example Let f(x) = arctan √ x. Find f′ (x). Solution d dx arctan √ x = 1 1 + (√ x )2 d dx √ x = 1 1 + x · 1 2 √ x = 1 2 √ x + 2x √ x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 26 / 32
  65. 65. . . . . . . Derivation: The derivative of arcsec Try this first. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  66. 66. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  67. 67. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  68. 68. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  69. 69. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . .x .1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  70. 70. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: . .x .1 . .y = arcsec x V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  71. 71. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: tan(arcsec x) = √ x2 − 1 1 . .x .1 . .y = arcsec x . √ x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  72. 72. . . . . . . Derivation: The derivative of arcsec Try this first. Let y = arcsec x, so x = sec y. Then sec y tan y dy dx = 1 =⇒ dy dx = 1 sec y tan y = 1 x tan(arcsec(x)) To simplify, look at a right triangle: tan(arcsec x) = √ x2 − 1 1 So d dx arcsec(x) = 1 x √ x2 − 1 . .x .1 . .y = arcsec x . √ x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 27 / 32
  73. 73. . . . . . . Another Example Example Let f(x) = earcsec 3x . Find f′ (x). V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
  74. 74. . . . . . . Another Example Example Let f(x) = earcsec 3x . Find f′ (x). Solution f′ (x) = earcsec 3x · 1 3x √ (3x)2 − 1 · 3 = 3earcsec 3x 3x √ 9x2 − 1 V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 28 / 32
  75. 75. . . . . . . Outline Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions Arcsine Arccosine Arctangent Arcsecant Applications V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 29 / 32
  76. 76. . . . . . . Application Example One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter stands 2 ft away from home plate as a pitch is thrown with a velocity of 130 ft/sec (about 90 mph). At what rate does the batter’s angle of gaze need to change to follow the ball as it crosses home plate? V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
  77. 77. . . . . . . Application Example One of the guiding principles of most sports is to “keep your eye on the ball.” In baseball, a batter stands 2 ft away from home plate as a pitch is thrown with a velocity of 130 ft/sec (about 90 mph). At what rate does the batter’s angle of gaze need to change to follow the ball as it crosses home plate? Solution Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 30 / 32
  78. 78. . . . . . . Solution Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
  79. 79. . . . . . . Solution Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ dt = 1 1 + (y/2)2 · 1 2 dy dt . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
  80. 80. . . . . . . Solution Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ dt = 1 1 + (y/2)2 · 1 2 dy dt When y = 0 and y′ = −130, then dθ dt y=0 = 1 1 + 0 · 1 2 (−130) = −65 rad/sec . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
  81. 81. . . . . . . Solution Let y(t) be the distance from the ball to home plate, and θ the angle the batter’s eyes make with home plate while following the ball. We know y′ = −130 and we want θ′ at the moment that y = 0. We have θ = arctan(y/2). Thus dθ dt = 1 1 + (y/2)2 · 1 2 dy dt When y = 0 and y′ = −130, then dθ dt y=0 = 1 1 + 0 · 1 2 (−130) = −65 rad/sec The human eye can only track at 3 rad/sec! . .2 ft .y .130 ft/sec . .θ V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 31 / 32
  82. 82. . . . . . . Summary y y′ arcsin x 1 √ 1 − x2 arccos x − 1 √ 1 − x2 arctan x 1 1 + x2 arccot x − 1 1 + x2 arcsec x 1 x √ x2 − 1 arccsc x − 1 x √ x2 − 1 Remarkable that the derivatives of these transcendental functions are algebraic (or even rational!) V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 32 / 32

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