lesson 17: inverse trigonometric functions
DESCRIPTION
The inverse trig functions are transcendental, but their derivatives are algebraic!TRANSCRIPT
. . . . . .
Section 3.5Inverse Trigonometric
FunctionsV63.0121, Calculus I
March 11–12, 2009
Announcements
I Get half of your unearned ALEKS points back by March 22
. . . . . .
What functions are invertible?
In order for f−1 to be a function, there must be only one a in Dcorresponding to each b in E.
I Such a function is called one-to-oneI The graph of such a function passes the horizontal line test:
any horizontal line intersects the graph in exactly one point if atall.
I If f is continuous, then f−1 is continuous.
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
. . . . . .
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.arcsin
I The domain of arcsin is [−1, 1]
I The range of arcsin is[−π
2,π
2
]
. . . . . .
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.arcsin
I The domain of arcsin is [−1, 1]
I The range of arcsin is[−π
2,π
2
]
. . . . . .
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
.arcsin
I The domain of arcsin is [−1, 1]
I The range of arcsin is[−π
2,π
2
]
. . . . . .
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
.arcsin
I The domain of arcsin is [−1, 1]
I The range of arcsin is[−π
2,π
2
]
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos..0
..π
.y = x
.arccos
I The domain of arccos is [−1, 1]
I The range of arccos is [0, π]
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos..0
..π
.y = x
.arccos
I The domain of arccos is [−1, 1]
I The range of arccos is [0, π]
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos..0
..π
.y = x
.arccos
I The domain of arccos is [−1, 1]
I The range of arccos is [0, π]
. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos..0
..π
.y = x
.arccos
I The domain of arccos is [−1, 1]
I The range of arccos is [0, π]
. . . . . .
arctanArctan 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
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, lim
x→−∞arctan x = −π
2
. . . . . .
arctanArctan 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
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, lim
x→−∞arctan x = −π
2
. . . . . .
arctanArctan 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
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, lim
x→−∞arctan x = −π
2
. . . . . .
arctanArctan 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
.arctan
.−π
2
.π
2
I The domain of arctan is (−∞,∞)
I The range of arctan is(−π
2,π
2
)I lim
x→∞arctan x =
π
2, lim
x→−∞arctan x = −π
2
. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant
. . . . . .
Theorem (The Inverse Function Theorem)Let f be differentiable at a, and f′(a) ̸= 0. Then f−1 is defined in an openinterval containing b = f(a), and
(f−1)′(b) =1
f′(f−1(b))
“Proof”.If y = f−1(x), then
f(y) = x,
So by implicit differentiation
f′(y)dydx
= 1 =⇒ dydx
=1
f′(y)=
1
f′(f−1(x))
. . . . . .
Theorem (The Inverse Function Theorem)Let f be differentiable at a, and f′(a) ̸= 0. Then f−1 is defined in an openinterval containing b = f(a), and
(f−1)′(b) =1
f′(f−1(b))
“Proof”.If y = f−1(x), then
f(y) = x,
So by implicit differentiation
f′(y)dydx
= 1 =⇒ dydx
=1
f′(y)=
1
f′(f−1(x))
. . . . . .
The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√
1 − x2
So
ddx
arcsin(x) =1√
1 − x2 .
.1.x
..y = arcsin x
.√
1 − x2
. . . . . .
The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√
1 − x2
So
ddx
arcsin(x) =1√
1 − x2
.
.1.x
..y = arcsin x
.√
1 − x2
. . . . . .
The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√
1 − x2
So
ddx
arcsin(x) =1√
1 − x2
.
.1.x
..y = arcsin x
.√
1 − x2
. . . . . .
The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√
1 − x2
So
ddx
arcsin(x) =1√
1 − x2
.
.1.x
..y = arcsin x
.√
1 − x2
. . . . . .
The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√
1 − x2
So
ddx
arcsin(x) =1√
1 − x2
.
.1.x
..y = arcsin x
.√
1 − x2
. . . . . .
The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√
1 − x2
So
ddx
arcsin(x) =1√
1 − x2
.
.1.x
..y = arcsin x
.√
1 − x2
. . . . . .
The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos ydydx
= 1 =⇒ dydx
=1
cos y=
1cos(arcsin x)
To simplify, look at a righttriangle:
cos(arcsin x) =√
1 − x2
So
ddx
arcsin(x) =1√
1 − x2 .
.1.x
..y = arcsin x
.√
1 − x2
. . . . . .
Graphing arcsin and its derivative
..|.−1
.|.1
.arcsin
.1√
1 − x2
. . . . . .
The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin ydydx
= 1 =⇒ dydx
=1
− sin y=
1− sin(arccos x)
To simplify, look at a righttriangle:
sin(arccos x) =√
1 − x2
So
ddx
arccos(x) = − 1√1 − x2 .
.1.√
1 − x2
.x.
.y = arccos x
. . . . . .
The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin ydydx
= 1 =⇒ dydx
=1
− sin y=
1− sin(arccos x)
To simplify, look at a righttriangle:
sin(arccos x) =√
1 − x2
So
ddx
arccos(x) = − 1√1 − x2 .
.1.√
1 − x2
.x.
.y = arccos x
. . . . . .
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 theirderivatives are opposites.
. . . . . .
Graphing arcsin and arccos
..|.−1
.|.1
.arcsin
.arccosNote
cos θ = sin(π
2− θ
)=⇒ arccos x =
π
2− arcsin x
So it’s not a surprise that theirderivatives are opposites.
. . . . . .
The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1 + x2
So
ddx
arctan(x) =1
1 + x2 .
.x
.1.
.y = arctan x
.√
1 + x2
. . . . . .
The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1 + x2
So
ddx
arctan(x) =1
1 + x2
.
.x
.1.
.y = arctan x
.√
1 + x2
. . . . . .
The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1 + x2
So
ddx
arctan(x) =1
1 + x2
.
.x
.1
..y = arctan x
.√
1 + x2
. . . . . .
The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1 + x2
So
ddx
arctan(x) =1
1 + x2
.
.x
.1.
.y = arctan x
.√
1 + x2
. . . . . .
The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1 + x2
So
ddx
arctan(x) =1
1 + x2
.
.x
.1.
.y = arctan x
.√
1 + x2
. . . . . .
The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1 + x2
So
ddx
arctan(x) =1
1 + x2
.
.x
.1.
.y = arctan x
.√
1 + x2
. . . . . .
The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2 ydydx
= 1 =⇒ dydx
=1
sec2 y= cos2(arctan x)
To simplify, look at a righttriangle:
cos(arctan x) =1√
1 + x2
So
ddx
arctan(x) =1
1 + x2 .
.x
.1.
.y = arctan x
.√
1 + x2
. . . . . .
Graphing arctan and its derivative
. .x
.y
.arctan
.1
1 + x2
. . . . . .
ExampleLet f(x) = arctan
√x. Find f′(x).
Solution
ddx
arctan√
x =1
1 +(√
x)2
ddx
√x =
11 + x
· 12√
x
=1
2√
x + 2x√
x
. . . . . .
ExampleLet f(x) = arctan
√x. Find f′(x).
Solution
ddx
arctan√
x =1
1 +(√
x)2
ddx
√x =
11 + x
· 12√
x
=1
2√
x + 2x√
x
. . . . . .
Recap
y y′
arcsin x1√
1 − x2
arccos x − 1√1 − x2
arctan x1
1 + x2
arccot x − 11 + x2
arcsec x1
x√
x2 − 1
arccsc x − 1
x√
x2 − 1
I Remarkable that thederivatives of thesetranscendental functionsare algebraic (or evenrational!)