1. The document discusses inverse trigonometric functions such as arcsin, arccos, and arctan. 2. It derives the derivatives of these inverse functions using the Inverse Function Theorem and properties of trigonometric functions. 3. The derivatives are derived to be 1/(√(1-x^2)) for arcsin, 1/√(1-x^2) for arccos, and 1/(1+x^2) for arctan.

1. Section 3.5
Inverse Trigonometric
Functions
V63.0121, Calculus I
March 11–12, 2009
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. . . . . .

2. 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.
. . . . . .

3. Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arctangent
Arcsecant
. . . . . .

4. arcsin
Arcsin is the inverse of the sine function after restriction to
[−π/2, π/2].
y
.
. . . x
.
s
. in
π π
− −
. .
2 2
. . . . . .

5. arcsin
Arcsin is the inverse of the sine function after restriction to
[−π/2, π/2].
y
.
. . . x
.
s
. in
π π
− −
. .
2 2
. . . . . .

6. arcsin
Arcsin is the inverse of the sine function after restriction to
[−π/2, π/2].
y
.
. =x
y
. . . x
.
s
. in
π π
− −
. .
2 2
. . . . . .

7. arcsin
Arcsin is the inverse of the sine function after restriction to
[−π/2, π/2].
y
.
a
. rcsin
. . . x
.
s
. in
π π
− −
. .
2 2
The domain of arcsin is [−1, 1]
[ π π]
The range of arcsin is − ,
22
. . . . . .

8. arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
y
.
c
. os
. . x
.
π
.
0
.
. . . . . .

9. arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
y
.
c
. os
. . x
.
π
.
0
.
. . . . . .

10. arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
y
.
. =x
y
c
. os
. . x
.
π
.
0
.
. . . . . .

11. arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
a
. rccos
y
.
c
. os
. . x
.
π
.
0
.
The domain of arccos is [−1, 1]
The range of arccos is [0, π]
. . . . . .

12. arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
y
.
. x
.
π π
3π 3π
−
− . .
. .
2 2
2 2
t
. an
. . . . . .

13. arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
y
.
. x
.
π π
3π 3π
−
− . .
. .
2 2
2 2
t
. an
. . . . . .

14. arctan
Arctan is the inverse of the tangent function after restriction to
. =x
y
[−π/2, π/2].
y
.
. x
.
π π
3π 3π
−
− . .
. .
2 2
2 2
t
. an
. . . . . .

15. arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
y
.
π
. a
. rctan
2
. x
.
π
−
.
2
The domain of arctan is (−∞, ∞)
( π π)
The range of arctan is − ,
22
π π
lim arctan x = , lim arctan x = −
2 x→−∞ 2
x→∞
. . . . . .

16. Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arctangent
Arcsecant
. . . . . .

17. 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
1
(f−1 )′ (b) = ′ −1
f (f (b))
. . . . . .

18. 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
1
(f−1 )′ (b) = ′ −1
f (f (b))
“Proof”.
If y = f−1 (x), then
f(y) = x,
So by implicit differentiation
dy dy 1 1
f′ (y) = 1 =⇒ =′ = ′ −1
dx dx f (y) f (f (x))
. . . . . .

19. The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
dy dy 1 1
= 1 =⇒ = =
cos y
dx dx cos y cos(arcsin x)
. . . . . .

20. The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
dy dy 1 1
= 1 =⇒ = =
cos y
dx dx cos y cos(arcsin x)
To simplify, look at a right
triangle:
.
. . . . . .

21. The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
dy dy 1 1
= 1 =⇒ = =
cos y
dx dx cos y cos(arcsin x)
To simplify, look at a right
triangle:
1
.
x
.
.
. . . . . .

22. The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
dy dy 1 1
= 1 =⇒ = =
cos y
dx dx cos y cos(arcsin x)
To simplify, look at a right
triangle:
1
.
x
.
. = arcsin x
y
.
. . . . . .

23. The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
dy dy 1 1
= 1 =⇒ = =
cos y
dx dx cos y cos(arcsin x)
To simplify, look at a right
triangle:
1
.
x
.
. = arcsin x
y
.√
. 1 − x2
. . . . . .

24. The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
dy dy 1 1
= 1 =⇒ = =
cos y
dx dx cos y cos(arcsin x)
To simplify, look at a right
triangle:
√
cos(arcsin x) = 1 − x2 1
.
x
.
. = arcsin x
y
.√
. 1 − x2
. . . . . .

25. The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
dy dy 1 1
= 1 =⇒ = =
cos y
dx dx cos y cos(arcsin x)
To simplify, look at a right
triangle:
√
cos(arcsin x) = 1 − x2 1
.
x
.
So
d 1
arcsin(x) = √ . = arcsin x
y
1 − x2 .√
dx
. 1 − x2
. . . . . .

26. Graphing arcsin and its derivative
1
.√
1 − x2
a
. rcsin
.
| |
. .
−
.1 1
.
. . . . . .

27. The derivative of arccos
Let y = arccos x, so x = cos y. Then
dy dy 1 1
− sin y = 1 =⇒ = =
− sin y − sin(arccos x)
dx dx
. . . . . .

28. The derivative of arccos
Let y = arccos x, so x = cos y. Then
dy dy 1 1
− sin y = 1 =⇒ = =
− sin y − sin(arccos x)
dx dx
To simplify, look at a right
triangle:
√
sin(arccos x) = 1 − x2 √
1
.
. 1 − x2
So
d 1 . = arccos x
y
arccos(x) = − √ .
1 − x2
dx x
.
. . . . . .

29. Graphing arcsin and arccos
a
. rccos
a
. rcsin
.
| |
. .
−
.1 1
.
. . . . . .

30. Graphing arcsin and arccos
a
. rccos
Note
(π )
−θ
cos θ = sin
2
a
. rcsin
π
=⇒ arccos x = − arcsin x
2
. So it’s not a surprise that their
| |
. .
−
.1 1
. derivatives are opposites.
. . . . . .

31. The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = cos2 (arctan x)
= 1 =⇒ =
sec2 y
dx dx
. . . . . .

32. The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = cos2 (arctan x)
= 1 =⇒ =
sec2 y
dx dx
To simplify, look at a right
triangle:
.
. . . . . .

33. The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = cos2 (arctan x)
= 1 =⇒ =
sec2 y
dx dx
To simplify, look at a right
triangle:
x
.
.
1
.
. . . . . .

34. The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = cos2 (arctan x)
= 1 =⇒ =
sec2 y
dx dx
To simplify, look at a right
triangle:
x
.
. = arctan x
y
.
1
.
. . . . . .

35. The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = cos2 (arctan x)
= 1 =⇒ =
sec2 y
dx dx
To simplify, look at a right
triangle:
√
x
.
. 1 + x2
. = arctan x
y
.
1
.
. . . . . .

36. The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = cos2 (arctan x)
= 1 =⇒ =
sec2 y
dx dx
To simplify, look at a right
triangle:
1
cos(arctan x) = √
√
1 + x2
x
.
. 1 + x2
. = arctan x
y
.
1
.
. . . . . .

37. The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = cos2 (arctan x)
= 1 =⇒ =
sec2 y
dx dx
To simplify, look at a right
triangle:
1
cos(arctan x) = √
√
1 + x2
x
.
. 1 + x2
So
d 1
. = arctan x
y
arctan(x) =
.
1 + x2
dx
1
.
. . . . . .

38. Graphing arctan and its derivative
y
.
a
. rctan
1
. x
.
1 + x2
. . . . . .

39. Example
√
x. Find f′ (x).
Let f(x) = arctan
. . . . . .

40. Example
√
x. Find f′ (x).
Let f(x) = arctan
Solution
√ d√
d 1 1 1
·√
(√ )2
arctan x = x=
1+x 2 x
dx x dx
1+
1
=√ √
2 x + 2x x
. . . . . .

41. Recap
y′
y
1
√
arcsin x
1 − x2
1 Remarkable that the
arccos x − √
1 − x2 derivatives of these
1 transcendental functions
arctan x
1 + x2 are algebraic (or even
1
−
arccot x rational!)
1 + x2
1
√
arcsec x
x x2 − 1
1
arccsc x − √
x x2 − 1
. . . . . .