CHAPTER 25 Derivatives of Inverse Trig Functions

[Pages:8]CHAPTER 25

Derivatives of Inverse Trig Functions

Our goal is simple, and the answers will come quickly. We will derive six new derivative formulas for the six inverse trigonometric functions:

h

i

d

?

1

sin (x)

dx

h

i

d

?

1

cos (x)

dx

h

i

d

?

1

tan (x)

dx

h

i

d

?

1

cot (x)

dx

h

i

d

?

1

sec (x)

dx

h

i

d

?

1

csc (x)

dx

These formulas will flow from the inverse rule from Chapter 24 (page 278):

h

i

d?

1

=

f (x)

0?

1

?

? .

1

dx

f f (x)

(25.1)

25.1 Derivatives of Inverse Sine and Cosine

Applying the inverse rule (25.1) with f (x) = sin(x) yields

h

i

d

? 1

=

sin (x)

?

1

?

?.

1

dx

cos sin (x)

(25.2)

Wcose?sairne?1a(xlm)? oinstththeedreen. omWienajtuosrt.

have to simplify To do this recall

the

!

? 1

=

the angle ? ? ? ?

2

2

sin (x) for which sin(?) = x .

Thus

? 1

sin (x)

It

is

the

angle

(between

??

and

?)

of

a

the

2

2

triangle on cause sin of

the this

uannigtlecierqclueawlshxo.)seTohpepnocsoiste?ssinid?e1(ixs)?xi.s(tBhee-

length of this side

the adjacpent side. By the

length is

? 1

2

x

.

Putting

Pyt?hag?o1rean? =thpeor?em2 cos sin (x) 1 x

into the above Equation (25.2), we get or latest rule:

1

?1

x

sin (x)

|

?

{z

?

?}

1

cos sin (x)

h

i

Rule 20

d

?

1

sin (x)

dx

= p1 ?2

1x

Derivatives of Inverse Sine and Cosine

287

We

reviewed

? 1

sin (x)

In

Section

6.1

and

presented

its

graph

on

page

101.

Figure 25.1 repeats the graph, along with the derivative from Rule 20.

=

? 1

y

f (x) cos (x)

? 2

1

-1

0 =p 1

f (x)

?2

1x

=

? 1

f (x) sin (x)

x 1

??

2

Figure 25.1.

The

function

? 1

sin (x)

and

its

derivative.

The derivativeis

always

positive,

reflecting

the

fact

that

the

tangents

to

? 1

sin

(x)

have

positive

slope.

The

derivative

has

vertical

asymptotes

at

= x

?1,

as

the

tangents

to

? 1

sin (x)

become

increasingly

steep

as

x

approaches

?1.

Now

consider

? 1

cos

(

x).

The

tangents to

its graph (Figure 25.2

below)

have negative slope, and the geometry suggests that its derivative is negative

the derivative of sin?1(x). Indeed this turns out to be exactly the case. This

chapter's Exercise 1 asks you to prove our next rule:

Rule 21

h

i

d

? 1

cos (x)

dx

=

? p1

?2 1x

=

? 1

?

f (x) cos (x)

-1 ?

x 1

1

0

? =p 1

f (x)

?2

1x

Figure

25.2.

The

function

? 1

cos (x)

and

its

derivative.

288

Derivatives of Inverse Trig Functions

25.2 Derivatives of Inverse Tangent and Cotangent

Now

let's

find

the

derivative

of

? 1

tan

(x).

Putting

f

(x)

=

tan(x)

into

the

inverse

rule

(25.1),

we

have

? 1

=

? 1

f (x) tan (x)

and

f

0

(x)

=

2

sec (

x),

and

we

get

h

i

d

?

1

tan (x)

dx

=

?

1

?

?

2

1

sec tan (x)

=

?

?

1

?

1

?? .

2

sec tan (x)

(25.3)

The

expression

sec

?? ?

1

tan (x)

in

the

denominator

is

the

length of the hypotenuse of the triangle to the right.

(See example Ppythagorean

6.3 in Chapter 6, page theorem, the length is

114? .)

By

? 1

t?he =

sec tan (x)

1

+

2

x

.

Inserting this into the above Equation (25.4)

yields

h

i

d dx

? 1

tan (x)

=

?

?

1

?

1

??

2

sec tan (x)

=

?p 1 ?

+

2 2

1x

=

1

+

.

2

1x

?

?

?1

) (x

n

ta

sec ?

x

1

tan (x)

1

We now have:

h

i

Rule 22

d

? 1

=1

tan (x)

+2

dx

1x

We

discussed

? 1

tan (x)

in

Chapter

6,

and

its

graph

is

in

Figure

6.3.

Below

Figure 25.3 repeats the graph, along with the derivative

1 2+

.

x1

?

=

? 1

f (x) tan (x)

2

1

x

0 =1

f (x) + 2

??

1x

2

Figure 25.3.

and

!lim?1

x

1 +2 1x

The = 0,

function

? 1

tan (

reflecting the

x) and its fact that

derivative

1 +

.

2

Note

1x

the tangent lines to

l!im1

x

=

1 +

2

=

1x

? 1

0

y tan (x)

become closer and closer to horizontal as ! ?1. The derivative bumps up

x

to 1 at = , where the tangent to =

? 1

is steepest, with slope 1

x0

y tan (x)

Exercise 3 below asks you to mirror the above arguments to deduce:

Rule 23

h d

? 1

i =

? 1

cot (x)

+2

dx

1x

Derivatives of Inverse Secant and Cosecant

289

25.3 Derivatives of Inverse Secant and Cosecant

We

reviewed

? 1

sec (x)

in

Section

6.3.

For

its

derivative,

put

= f (x) sec(x)

into

the

inverse

rule

(25.1),

with

? 1

=

? 1

f (x) sec (x)

and

f 0(x) = sec(x) tan(x).

We

get

h

i

d

? 1

=

sec (x)

0?

1

?

?

1

dx

f f (x)

=

??

1

?1 ?

?? ?

1

sec f (x) tan f (x)

=

?

? 1

?1 ?

??

? .

1

sec sec (x) tan sec (x)

Because

sec

?? ?

1

sec (x)

=

x,

the

above

becomes

h

i

d

? 1

sec (x)

dx

= ?

?1 ?

? .

1

x tan sec (x)

(p

In

Example

6.5

we

showed

that

?? ?

1

tan sec (x)

=

px2 ? 1 ? 2?

x1

With this, Equation 25.4 above becomes

(25.4)

if x is positive if x is negative

8

h

i

d

?

1

sec (x)

dx

=

>>< >>:

p1 2?

xx 1 p1

? 2?

xx 1

if x is positive if x is negative.

But

if

x

is

negative,

then

? x

is

positive,

and

the

above

consolidates

to

h

i

Rule 24

d

?

1

sec (x)

dx

= p1 | | 2? xx 1

This

graph

of

? 1

sec (x)

and

its

derivative

is

shown

in

Figure

25.3.

y

?

? 2

? 11

=

=

? 1

y f (x) sec (x)

= 0 = p1 y f (x) | | 2 ?

xx 1

Figure 25.4.

The graph of

? 1

and its derivative.

sec (x)

The domain of

both

functions

is

?1 (,

? 1]

[

[1,

1).

Note that the derivative has vertical

asymptotes

at

x = ?1,

where

the

tangent

line

to

=

? 1

y sec (x)

is

vertical.

290

Derivatives of Inverse Trig Functions

This chapter's Exercise 2 asks you to use reasoning similar to the above to deduce our final rule.

Rule 25

h

i

d

?

1

csc (x)

dx

=

? p1 | | 2? xx 1

Each of our new rules has a chain rule generalization. For example, Rule 25 generalizes as

h d

??

?i

1

csc g(x)

dx

=

??

? ??q 1

2?

0

g (x)

=

??

?0 ??qg (x) 2 ? .

g(x) (g(x)) 1

g(x) (g(x)) 1

Here is a summary of this Chapter's new rules, along with their chain rule generalizations.

h

i

d

?

1

= p1

sin (x) dx

?2 1x

h d

? 1

i

?

= p1

cos (x) dx

?2 1x

h

i

d

? 1

=1

tan (x)

+2

dx

1x

h

i

d

?

1

=

? 1

cot (x)

+2

dx

1x

h

i

d

?

1

= p1

sec (x) dx

| | 2? xx 1

h

i

d

? 1

csc (x)

dx

=

? p1 | | 2? xx 1

h d

??

?i

1

sin g(x)

=

q

1

0

g (x)

dx

?

2

h d

??

?i

1

cos g(x)

=

1 (g(x))

? q1

0

g (x)

dx

?

2

h d

??

1

?i =

1 (g(x))

1

0

tan g(x) dx

+

g (x)

2

1 (g(x))

h d

??

1

?i =

? 1

0

cot g(x) dx

+

g (x)

2

1 (g(x))

h d

??

?i

1

sec g(x)

dx

=

??

0

??qg (x) 2 ?

h d

??

1

?i

csc g(x)

dx

=

g(x) (g(x)) 1

?0

??

??qg (x) 2 ?

g(x) (g(x)) 1

Example 25.1

hp d

? 1

i

h?

=d

? 1

?1 i ? 2 =1

?

?? 1

h d

1

2

i

? 1

cos (x)

cos (x)

cos (x)

cos (x)

dx

dx

2

dx

? =1

? 1

?? 1

2

? p1

=

cos (x) 2

?2 1x

?

p

1p

.

? 1

?2

2 cos (x) 1 x

h

i

h

i

? 1

Example 25.2

?

?

d

1

tan (x)

=

1d

tan (x)

? 1

?

=

1

tan (x)

1

tan (x)

=e

.

e

e

tan (x) e

+2

+2

dx

dx

1x 1x

Example 25.3

h d

??

1

?i x=

1

hi d x = 1 x=

tan e dx

+ x2 1 (e )

dx

e

+

e

2x

1e

x

e +

.

2x

1e

Summary of Derivative Rules

291

25.4 Summary of Derivative Rules

We have reached the end of our derivative rules. In summary, we have the following rules for specific functions. The corresponding chain rule generalizations are shown to the right.

Constant Rule Identity Rule Power Rule Exp Rules

Log Rules

Trig Rules

Inverse Trig Rules

Rule

?? d=

c0

dx ??

d= x1

dxh i

d

n=

? n1

x nx

dxh i

d x =x ee

dxh i

d x=

x

a ln(a)a

dxh

?

d

=1

ln(x)

dxh

ix

d

=1

loga(x)

dxh

i x ln(a)

d

=

sin(x) cos(x)

dxh

i

d

=?

cos(x) sin(x)

dxh

i

d

=

2

tan(x) sec (x)

dxh

i

d

=? 2

cot(x) csc (x)

dxh

i

d

=

sec(x) sec(x) tan(x)

dxh

i

d

=?

csc(x) csc(x) cot(x)

dxh

i

d

? 1

=p 1

sin (x)

dx

h

i

d

?

1

=

?2 1x ? p1

cos (x)

dx

h

i

?2 1x

d

? 1

=1

tan (x)

+2

dxh d

? 1

i =

1x ?

1

cot (x)

+2

dxh

i 1x

d dx d dx

? 1

sec (x)

h

i

? 1

csc (x)

= =

p1 || 2 xx

? p1 || 2 xx

? 1

? 1

Chain Rule Generalization

?? d

?? ? n=

??

n1

0

g(x) n g(x) g (x)

dxh i

d

g(x)

=

0 g(x)

e

e g (x)

dxh i

d g(x) =

0 g(x)

a

ln(a)a g (x)

dxh ? d

?i =1

0

ln g(x)

g (x)

dx h ? ? i g(x)

d

=1

loga g(x)

0

g (x)

dxh ? d

?i =

g(x) ln(a) ? ?0

sin g(x) cos g(x) g (x)

dxh ? d

?i

?

=?

?0

cos g(x)

sin g(x) g (x)

dx h ? ?i

d

=

? ?0

2

tan g(x) sec g(x) g (x)

dxh ? d

?i

?

=? 2

?0

cot g(x)

csc g(x) g (x)

dx h ? ?i

d

=

? ? ? ?0

sec g(x) sec g(x) tan g(x) g (x)

dxh ? d

?i

?

=?

??

?0

csc g(x)

csc g(x) cot g(x) g (x)

dx d dx d dx d

h h h

? ? ?i

1

sin g(x)

? ? ?i

1

cos g(x)

? ? ?i

1

= = =

p 1

p 1

1

?

2

( g(x))

?

1

?

2

( g(x))

1

0

g (x)

0

g (x)

0

tan g(x)

+

g (x)

2

dxh d

??

1

?i =

1

( g( x)) ?

1

0

cot g(x)

+

g (x)

2

dx d dx d dx

h ? ? ?i

1

sec g(x)

h ? ? ?i

1

csc g(x)

= =

1 (g(x))

p1

||

2

g(x) (g(x))

?

p1

||

2

g(x) (g(x))

? 1

? 1

0

g (x)

0

g (x)

292

Derivatives of Inverse Trig Functions

In addition we have the following general rules for the derivatives of combinations of functions.

?

?

Constant Multiple Rule:

d c f (x)

=

0

c f (x)

Sum/Di erence Rule:

dx

? d

?

? =

0 ?0

f (x) g(x)

f (x) g (x)

Product Rule:

dx

?

?

d

f (x)g(x)

=0

+

0

f (x)g(x) f (x)g (x)

Quotient Rule:

dx d f (x)

0

?

0

= f (x)g(x) f (x)g (x)

dx g(x)

2

( g( x))

Chain Rule:

? ? ?? d

f g(x)

=0

0

f (g(x)) g (x)

Inverse Rule:

dx

? d

?

?

1

f (x)

dx

=

0?

1

?

?

1

f f (x)

We used this last rule, the inverse rule, to find the derivatives of ln(x) and the inverse trig functions. After it has served these purposes it is mostly

retired for the remainder of Calculus I, except for the stray exercise or quiz

or test question.

This looks like a lot of rules to remember, and it is. But through practice

and usage you will reach the point of using them automatically, with hardly

a thought. Be sure to get enough practice!

Exercises for Chapter 25

1. Show that

d dx

h

i

? 1

cos (x)

=

? p1

?2 1x

.

2. Show that

h d

? 1

i =

? p1

.

csc (x) dx

| | 2? xx 1

3. Show that

h

i

d

? 1

=

cot (x)

? 1

.

+2

dx

1x

Find the derivatives of the given functions.

4. 7.

? ?p ?

1

sin

? 1

2x ?? ?

tan x

5. 8.

???

1

ln

tan

? 1

? ?

(?x)

sec x

10.

? 1

? ?

?

cos x

11.

?? ?

15

sec x

13.

??

1

? +?

tan ln(x)

14.

??

?

1

tan x sin(x)

6. 9.

?

x

1

e

tan ??

(x) ?

1

ln sin (x)

12.

? 1

?

tan ( x)

e

15.

??

?

1

x sin ln(x)

Summary of Derivative Rules

293

Exercise Solutions for Chapter 25

1. Show that

d dx

h

i

? 1

=

cos (x)

h

? p1

?2 1 xi

.

By the inverse rule, d

? 1

cos (x)

dx

= ?

? 1 ? ?.

1

sin cos (x)

Now we simplify the denominator.

Fcpo1rs?o?xm12(=x)tphwee?gset2ta. nsWidnia?trchodst?hd1i(sixa,)gt?hr=aemHOaPYbPPofvo=er

1

1 hx

becomes d

? 1

i

?

=p 1

.

?

cos (x) dx

?2 1x

1

? 1

cos (x)

0

x

3. Show that

h

i

d

? 1

=

cot (x)

? 1

.

+2

dx

1x

Suggestion: Verify the identity

? 1

=??

? 1

. Then di erentiate

cot (x) tan (x)

2

both sides of this.

5.

h? d

?

?i

1

ln tan (x)

dx

=

1

? 1

tan (x)

h

i

d

?

1

tan (x)

dx

=

1

? 1

tan (x)

1 +2 1x

=

? 1

1?

+

tan (x) 1

?

2

x

7.

h d

? 1

? ?

?i =

tan x

? +?

=

2

? + ?2 2

dx

1 ( x) 1 x

9.

h? d

?

?i

1

ln sin (x)

dx

=

1

? 1

sin (x)

h

i

d

?

1

sin (x)

dx

=

1

? 1

sin (x)

p1 ?2

1x

=

? 1

1p ?2

sin (x) 1 x

11.

d dx

h

? 1

sec

??

5

x

i

=

??

5

??q?1

5

?

2

?

4

4

5x

=

??

5??p5x10 ?

= ??

??p

5

10

?

x x 1 xx 1

xx 1

13.

h d

??

1

?i +?

tan ln(x)

dx

=

?1 ?

+

2

1 ln(x)

1 x

=

?1 ?

+

2

x x ln(x)

15.

h d

??

1

?i =?

??

1

?

h

+d

??

1

?i

x sin ln(x) 1 sin ln(x) x sin ln(x)

dx

dx

=

??

1

sin

? ln(x)

+

x

q

1 ?

?

1

=

??

1

? +

sin ln(x)

q

1 ?

?

?

2x

?

2

1 ln(x)

1 ln(x)

 ................
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