. . . . . .

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!)