lesson 16: inverse trigonometric functions (slides)

..

Sec on 3.6Inverse Trigonometric Func ons

V63.0121.011: Calculus IProfessor Ma hew Leingang

New York University

March 28, 2011

AnnouncementsI Midterm has been returned. Please seeFAQ on Blackboard (under ”Exams andQuizzes”)

I Quiz 3 this week in recita on onSec on 2.6, 2.8, 3.1, 3.2

I Quiz 4 April 14–15 on Sec ons 3.3, 3.4,3.5, and 3.7

I Quiz 5 April 28–29 on Sec ons 4.1, 4.2,4.3, and 4.4

Objectives

I Know the defini ons, domains, ranges,and other proper es of the inversetrignometric func ons: arcsin, arccos,arctan, arcsec, arccsc, arccot.

I Know the deriva ves of the inversetrignometric func ons.

Outline

Inverse Trigonometric Func ons

Deriva ves of Inverse Trigonometric Func onsArcsineArccosineArctangentArcsecant

Applica ons

What is an inverse function?Defini onLet f be a func on with domain D and range E. The inverse of f is thefunc on f−1 defined by:

f−1(b) = a,

where a is chosen so that f(a) = b.

Sof−1(f(x)) = x, f(f−1(x)) = x

What is an inverse function?Defini onLet f be a func on with domain D and range E. The inverse of f is thefunc on f−1 defined by:

f−1(b) = a,

where a is chosen so that f(a) = b.

Sof−1(f(x)) = x, f(f−1(x)) = x

What functions are invertible?

In order for f−1 to be a func on, there must be only one a in Dcorresponding to each b in E.

I Such a func on is called one-to-oneI The graph of such a func on passes the horizontal line test:any horizontal line intersects the graph in exactly one point if atall.

I If f is con nuous, then f−1 is con nuous.

Graphing the inverse functionI If b = f(a), then f−1(b) = a.

I So if (a, b) is on the graph of f,then (b, a) is on the graph of f−1.

I On the xy-plane, the point (b, a)is the reflec on of (a, b) in theline y = x.

I Therefore:

..x

.

y

..

(a, b)

..

(b, a)

.

y = x

FactThe graph of f−1 is the reflec on of the graph of f in the line y = x.

Graphing the inverse functionI If b = f(a), then f−1(b) = a.I So if (a, b) is on the graph of f,then (b, a) is on the graph of f−1.

I On the xy-plane, the point (b, a)is the reflec on of (a, b) in theline y = x.

I Therefore:

..x

.

y

..

(a, b)

..

(b, a)

.

y = x

FactThe graph of f−1 is the reflec on of the graph of f in the line y = x.

Graphing the inverse functionI If b = f(a), then f−1(b) = a.I So if (a, b) is on the graph of f,then (b, a) is on the graph of f−1.

I On the xy-plane, the point (b, a)is the reflec on of (a, b) in theline y = x.

I Therefore:

..x

.

y

..

(a, b)

..

(b, a)

.

y = x

FactThe graph of f−1 is the reflec on of the graph of f in the line y = x.

Graphing the inverse functionI If b = f(a), then f−1(b) = a.I So if (a, b) is on the graph of f,then (b, a) is on the graph of f−1.

I On the xy-plane, the point (b, a)is the reflec on of (a, b) in theline y = x.

I Therefore:

..x

.

y

..

(a, b)

..

(b, a)

.

y = x

FactThe graph of f−1 is the reflec on of the graph of f in the line y = x.

Graphing the inverse functionI If b = f(a), then f−1(b) = a.I So if (a, b) is on the graph of f,then (b, a) is on the graph of f−1.

I On the xy-plane, the point (b, a)is the reflec on of (a, b) in theline y = x.

I Therefore:

..x

.

y

..

(a, b)

..

(b, a)

.

y = x

FactThe graph of f−1 is the reflec on of the graph of f in the line y = x.

Graphing the inverse functionI If b = f(a), then f−1(b) = a.I So if (a, b) is on the graph of f,then (b, a) is on the graph of f−1.

I On the xy-plane, the point (b, a)is the reflec on of (a, b) in theline y = x.

I Therefore:..

x.

y

..

(a, b)

..

(b, a)

.

y = x

FactThe graph of f−1 is the reflec on of the graph of f in the line y = x.

Graphing the inverse functionI If b = f(a), then f−1(b) = a.I So if (a, b) is on the graph of f,then (b, a) is on the graph of f−1.

I On the xy-plane, the point (b, a)is the reflec on of (a, b) in theline y = x.

I Therefore:..

x.

y

..

(a, b)

..

(b, a)

.

y = x

FactThe graph of f−1 is the reflec on of the graph of f in the line y = x.

arcsinArcsin is the inverse of the sine func on a er restric on 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

]

arcsinArcsin is the inverse of the sine func on a er restric on 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

]

arcsinArcsin is the inverse of the sine func on a er restric on 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

]

arcsinArcsin is the inverse of the sine func on a er restric on 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

]

arccosArccos is the inverse of the cosine func on a er restric on 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]

arccosArccos is the inverse of the cosine func on a er restric on 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]

arccosArccos is the inverse of the cosine func on a er restric on 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]

arccosArccos is the inverse of the cosine func on a er restric on 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 func on a er restric on to(−π/2, π/2).

.. x.

y

.

tan

.

−3π2

.−π

2

.π

2

.3π2

.

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, limx→−∞

arctan x = −π

2

arctanArctan is the inverse of the tangent func on a er restric on to(−π/2, π/2).

.. x.

y

.

tan

.

−3π2

.−π

2

.π

2

.3π2

.

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, limx→−∞

arctan x = −π

2

arctanArctan is the inverse of the tangent func on a er restric on to(−π/2, π/2).

.. x.

y

.

tan

.

−3π2

.−π

2

.π

2

.3π2

.

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, limx→−∞

arctan x = −π

2

arctanArctan is the inverse of the tangent func on a er restric on to(−π/2, π/2).

.. x.

y

.

tan

.

−3π2

.−π

2

.π

2

.3π2

.

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, limx→−∞

arctan x = −π

2

arcsecArcsecant is the inverse of secant a er restric on to[0, π/2) ∪ [π, 3π/2).

.. x.

y

.

sec

.

−3π2

.−π

2

.π

2

.3π2

.

y = x

...

π

2.

3π2

I The domain of arcsec is (−∞,−1] ∪ [1,∞)

I The range of arcsec is[0,

π

2

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

arcsecArcsecant is the inverse of secant a er restric on to[0, π/2) ∪ [π, 3π/2).

.. x.

y

.

sec

.

−3π2

.−π

2

.π

2

.3π2

..

.

y = x

...

π

2.

3π2

I The domain of arcsec is (−∞,−1] ∪ [1,∞)

I The range of arcsec is[0,

π

2

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

arcsecArcsecant is the inverse of secant a er restric on to[0, π/2) ∪ [π, 3π/2).

.. x.

y

.

sec

.

−3π2

.−π

2

.π

2

.3π2

...

y = x

...

π

2.

3π2

I The domain of arcsec is (−∞,−1] ∪ [1,∞)

I The range of arcsec is[0,

π

2

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

arcsecArcsecant is the inverse of secant a er restric on to[0, π/2) ∪ [π, 3π/2).

.. x.

y

.

sec

.

−3π2

.−π

2

.π

2

.3π2

..

.

y = x

...

π

2.

3π2

I The domain of arcsec is (−∞,−1] ∪ [1,∞)

I The range of arcsec is[0,

π

2

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

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

Check: Values of inverse trigonometric functions

Example

FindI arcsin(1/2)I arctan(−1)

I arccos

(−√22

)

Solu on

Iπ

6I −π

4I

3π4

Check: Values of inverse trigonometric functions

Example

FindI arcsin(1/2)I arctan(−1)

I arccos

(−√22

)

Solu on

Iπ

6

I −π

4I

3π4

What is arctan(−1)?

.....

3π/4

..

−π/4

.

sin(3π/4) =√22

.

cos(3π/4) = −√22

.

sin(π/4) = −√22

.cos(π/4) =

√22

I Yes, tan(3π4

)= −1

I But, the range of arctan is(−π

2,π

2

)I Another angle whosetangent is−1 is−π

4, and

this is in the right range.I So arctan(−1) = −π

4

What is arctan(−1)?

.....

3π/4

..

−π/4

.

sin(3π/4) =√22

.

cos(3π/4) = −√22

.

sin(π/4) = −√22

.cos(π/4) =

√22

I Yes, tan(3π4

)= −1

I But, the range of arctan is(−π

2,π

2

)I Another angle whosetangent is−1 is−π

4, and

this is in the right range.I So arctan(−1) = −π

4

What is arctan(−1)?

.....

3π/4

..

−π/4

.

sin(3π/4) =√22

.

cos(3π/4) = −√22

.

sin(π/4) = −√22

.cos(π/4) =

√22

I Yes, tan(3π4

)= −1

I But, the range of arctan is(−π

2,π

2

)

I Another angle whosetangent is−1 is−π

4, and

this is in the right range.I So arctan(−1) = −π

4

What is arctan(−1)?

.....

3π/4

..

−π/4

.

sin(3π/4) =√22

.

cos(3π/4) = −√22

.

sin(π/4) = −√22

.cos(π/4) =

√22

I Yes, tan(3π4

)= −1

I But, the range of arctan is(−π

2,π

2

)I Another angle whosetangent is−1 is−π

4, and

this is in the right range.

I So arctan(−1) = −π

4

What is arctan(−1)?

.....

3π/4

..

−π/4

.

sin(3π/4) =√22

.

cos(3π/4) = −√22

.

sin(π/4) = −√22

.cos(π/4) =

√22

I Yes, tan(3π4

)= −1

I But, the range of arctan is(−π

2,π

2

)I Another angle whosetangent is−1 is−π

4, and

this is in the right range.I So arctan(−1) = −π

4

Check: Values of inverse trigonometric functions

Example

FindI arcsin(1/2)I arctan(−1)

I arccos

(−√22

)

Solu on

Iπ

6I −π

4

I3π4

Check: Values of inverse trigonometric functions

Example

FindI arcsin(1/2)I arctan(−1)

I arccos

(−√22

)

Solu on

Iπ

6I −π

4I

3π4

Caution: Notational ambiguity

..sin2 x = (sin x)2. sin−1 x = (sin x)−1

I sinn xmeans the nth power of sin x, except when n = −1!I The book uses sin−1 x for the inverse of sin x, and never for(sin x)−1.

I I use csc x for1

sin xand arcsin x for the inverse of sin x.

Outline

Inverse Trigonometric Func ons

Deriva ves of Inverse Trigonometric Func onsArcsineArccosineArctangentArcsecant

Applica ons

The Inverse Function TheoremTheorem (The Inverse Func on Theorem)

Let f be differen able at a, and f′(a) ̸= 0. Then f−1 is defined in anopen interval containing b = f(a), and

(f−1)′(b) =1

f′(f−1(b))

In Leibniz nota on we have

dxdy

=1

dy/dx

Illustrating the IFTExample

Use the inverse func on theorem to find the deriva ve of thesquare root func on.

Illustrating the IFTExample

Use the inverse func on theorem to find the deriva ve of thesquare root func on.

Solu on (Newtonian nota on)

Let f(x) = x2 so that f−1(y) =√y. Then f′(u) = 2u so for any b > 0

we have(f−1)′(b) =

12√b

Illustrating the IFTExample

Use the inverse func on theorem to find the deriva ve of thesquare root func on.

Solu on (Leibniz nota on)

If the original func on is y = x2, then the inverse func on is definedby x = y2. Differen ate implicitly:

1 = 2ydydx

=⇒ dydx

=1

2√x

The derivative of arcsineLet y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

=1

cos y=

1cos(arcsin x)

To simplify, look at a right triangle:

cos(arcsin x) =√

1− x2

SoFact

ddx

arcsin(x) =1√

1− x2.

.... y = arcsin x.

1

.

x

.√1− x2

The derivative of arcsineLet y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

=1

cos y=

1cos(arcsin x)

To simplify, look at a right triangle:

cos(arcsin x) =√

1− x2

SoFact

ddx

arcsin(x) =1√

1− x2

.

.... y = arcsin x.

1

.

x

.√1− x2

The derivative of arcsineLet y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

=1

cos y=

1cos(arcsin x)

To simplify, look at a right triangle:

cos(arcsin x) =√

1− x2

SoFact

ddx

arcsin(x) =1√

1− x2

..... y = arcsin x

.

1

.

x

.√1− x2

The derivative of arcsineLet y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

=1

cos y=

1cos(arcsin x)

To simplify, look at a right triangle:

cos(arcsin x) =√

1− x2

SoFact

ddx

arcsin(x) =1√

1− x2

..... y = arcsin x.

1

.

x

.√1− x2

The derivative of arcsineLet y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

=1

cos y=

1cos(arcsin x)

To simplify, look at a right triangle:

cos(arcsin x) =√

1− x2

SoFact

ddx

arcsin(x) =1√

1− x2

..... y = arcsin x.

1

.

x

.√1− x2

The derivative of arcsineLet y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

=1

cos y=

1cos(arcsin x)

To simplify, look at a right triangle:

cos(arcsin x) =√

1− x2

SoFact

ddx

arcsin(x) =1√

1− x2

..... y = arcsin x.

1

.

x

.√1− x2

The derivative of arcsineLet y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

=1

cos y=

1cos(arcsin x)

To simplify, look at a right triangle:

cos(arcsin x) =√

1− x2

SoFact

ddx

arcsin(x) =1√

1− x2..... y = arcsin x.

1

.

x

.√1− x2

Graphing arcsin and its derivative

I The domain of f is [−1, 1],but the domain of f′ is(−1, 1)

I limx→1−

f′(x) = +∞

I limx→−1+

f′(x) = +∞ ..| .−1

. |.1

...

arcsin

.

1√1− x2

Composing with arcsinExample

Let f(x) = arcsin(x3 + 1). Find f′(x).

Solu onWe have

ddx

arcsin(x3 + 1) =1√

1− (x3 + 1)2ddx

(x3 + 1)

=3x2√

−x6 − 2x3

Composing with arcsinExample

Let f(x) = arcsin(x3 + 1). Find f′(x).

Solu onWe have

ddx

arcsin(x3 + 1) =1√

1− (x3 + 1)2ddx

(x3 + 1)

=3x2√

−x6 − 2x3

The derivative of arccosLet y = arccos x, so x = cos y. Then

− sin ydydx

= 1 =⇒ dydx

=1

− sin y=

1− sin(arccos x)

To simplify, look at a right triangle:

sin(arccos x) =√

1− x2

SoFact

ddx

arccos(x) = − 1√1− x2

..

1

.

√1− x2

.x

.. y = arccos x

The derivative of arccosLet y = arccos x, so x = cos y. Then

− sin ydydx

= 1 =⇒ dydx

=1

− sin y=

1− sin(arccos x)

To simplify, look at a right triangle:

sin(arccos x) =√

1− x2

SoFact

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(π2− θ)

=⇒ arccos x =π

2− arcsin x

So it’s not a surprise that theirderiva ves are opposites.

Graphing arcsin and arccos

..| .−1

. |.1

...

arcsin

...

arccos Note

cos θ = sin(π2− θ)

=⇒ arccos x =π

2− arcsin x

So it’s not a surprise that theirderiva ves are opposites.

The derivative of arctanLet y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

=1

sec2 y= cos2(arctan x)

To simplify, look at a right triangle:

cos(arctan x) =1√

1+ x2

SoFact

ddx

arctan(x) =1

1+ x2.

.... y = arctan x.

x

.1

.

√1+ x2

The derivative of arctanLet y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

=1

sec2 y= cos2(arctan x)

To simplify, look at a right triangle:

cos(arctan x) =1√

1+ x2

SoFact

ddx

arctan(x) =1

1+ x2

.

.... y = arctan x.

x

.1

.

√1+ x2

The derivative of arctanLet y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

=1

sec2 y= cos2(arctan x)

To simplify, look at a right triangle:

cos(arctan x) =1√

1+ x2

SoFact

ddx

arctan(x) =1

1+ x2

..... y = arctan x

.

x

.1

.

√1+ x2

The derivative of arctanLet y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

=1

sec2 y= cos2(arctan x)

To simplify, look at a right triangle:

cos(arctan x) =1√

1+ x2

SoFact

ddx

arctan(x) =1

1+ x2

..... y = arctan x.

x

.1

.

√1+ x2

The derivative of arctanLet y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

=1

sec2 y= cos2(arctan x)

To simplify, look at a right triangle:

cos(arctan x) =1√

1+ x2

SoFact

ddx

arctan(x) =1

1+ x2

..... y = arctan x.

x

.1

.

√1+ x2

The derivative of arctanLet y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

=1

sec2 y= cos2(arctan x)

To simplify, look at a right triangle:

cos(arctan x) =1√

1+ x2

SoFact

ddx

arctan(x) =1

1+ x2

..... y = arctan x.

x

.1

.

√1+ x2

The derivative of arctanLet y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

=1

sec2 y= cos2(arctan x)

To simplify, look at a right triangle:

cos(arctan x) =1√

1+ x2

SoFact

ddx

arctan(x) =1

1+ x2..... y = arctan x.

x

.1

.

√1+ x2

Graphing arctan and its derivative

.. x.

y

.

arctan

.

11+ x2.

π/2

.

−π/2

I The domain of f and f′ are both (−∞,∞)I Because of the horizontal asymptotes, lim

x→±∞f′(x) = 0

Composing with arctanExample

Let f(x) = arctan√x. Find f′(x).

Solu on

ddx

arctan√x =

1

1+(√

x)2 d

dx√x =

11+ x

· 12√x

=1

2√x+ 2x

√x

Composing with arctanExample

Let f(x) = arctan√x. Find f′(x).

Solu on

ddx

arctan√x =

1

1+(√

x)2 d

dx√x =

11+ x

· 12√x

=1

2√x+ 2x

√x

The derivative of arcsecTry this first.

Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1 .

.

x

.1

.. y = arcsec x.

√x2 − 1

The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1 .

.

x

.1

.. y = arcsec x.

√x2 − 1

The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1

.

.

x

.1

.. y = arcsec x.

√x2 − 1

The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1

.

.

x

.1

.. y = arcsec x.

√x2 − 1

The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1

.

.

x

.1

.. y = arcsec x

.

√x2 − 1

The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1

..

x

.1

.. y = arcsec x

.

√x2 − 1

The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1

..

x

.1

.. y = arcsec x.

√x2 − 1

The derivative of arcsecTry this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

=1

sec y tan y=

1x tan(arcsec(x))

To simplify, look at a right triangle:

tan(arcsec x) =√x2 − 11

SoFact

ddx

arcsec(x) =1

x√x2 − 1 ..

x

.1

.. y = arcsec x.

√x2 − 1

Another ExampleExample

Let f(x) = earcsec 3x. Find f′(x).

Solu on

f′(x) = earcsec 3x · 13x√(3x)2 − 1

· 3

=3earcsec 3x

3x√9x2 − 1

Another ExampleExample

Let f(x) = earcsec 3x. Find f′(x).

Solu on

f′(x) = earcsec 3x · 13x√(3x)2 − 1

· 3

=3earcsec 3x

3x√9x2 − 1

Outline

Inverse Trigonometric Func ons

Deriva ves of Inverse Trigonometric Func onsArcsineArccosineArctangentArcsecant

Applica ons

ApplicationExampleOne of the guiding principles of mostsports is to “keep your eye on theball.” In baseball, a ba er stands 2 ftaway from home plate as a pitch isthrown with a velocity of 130 ft/sec(about 90mph). At what rate doesthe ba er’s angle of gaze need tochange to follow the ball as it crosseshome plate?

Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0.

We have θ = arctan(y/2). Thusdθdt

=1

1+ (y/2)2· 12dydt

When y = 0 and y′ = −130, then

dθdt

∣∣∣∣y=0

=1

1+ 0· 12(−130) = −65 rad/sec

The human eye can only track at 3 rad/sec!

..2

.

y

.

130 ft/sec

..θ

Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0.We have θ = arctan(y/2). Thus

dθdt

=1

1+ (y/2)2· 12dydt

When y = 0 and y′ = −130, then

dθdt

∣∣∣∣y=0

=1

1+ 0· 12(−130) = −65 rad/sec

The human eye can only track at 3 rad/sec!

..2

.

y

.

130 ft/sec

..θ

Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0.We have θ = arctan(y/2). Thus

dθdt

=1

1+ (y/2)2· 12dydt

When y = 0 and y′ = −130, then

dθdt

∣∣∣∣y=0

=1

1+ 0· 12(−130) = −65 rad/sec

The human eye can only track at 3 rad/sec!

..2

.

y

.

130 ft/sec

..θ

Solu onLet y(t) be the distance from the ball to home plate, and θ the anglethe ba er’s eyes make with home plate while following the ball. Weknow y′ = −130 and we want θ′ at the moment that y = 0.We have θ = arctan(y/2). Thus

dθdt

=1

1+ (y/2)2· 12dydt

When y = 0 and y′ = −130, then

dθdt

∣∣∣∣y=0

=1

1+ 0· 12(−130) = −65 rad/sec

The human eye can only track at 3 rad/sec!..

2.

y

.

130 ft/sec

..θ

Summary

y y′ y y′

arcsin x1√

1− x2arccos x − 1√

1− x2

arctan x1

1+ x2arccot x − 1

1+ x2

arcsec x1

x√x2 − 1

arccsc x − 1x√x2 − 1

I Remarkable that the deriva ves of these transcendentalfunc ons are algebraic (or even ra onal!)