We cover the inverses to the trigonometric functions sine, cosine, tangent, cotangent, secant, cosecant, and their derivatives. The remarkable fact is that although these functions and their inverses are transcendental (complicated) functions, the derivatives are algebraic functions. Also, we meet my all-time favorite function: arctan.
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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
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. . . . . . .
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. . . . . . .
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
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
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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. . . . . . .
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