Inverse trigonometric functions (Sect. 7.6) Review ...

Inverse trigonometric functions (Sect. 7.6)

Today: Derivatives and integrals.

Review: Definitions and properties. Derivatives. Integrals.

Last class: Definitions and properties.

Domains restrictions and inverse trigs. Evaluating inverse trigs at simple values. Few identities for inverse trigs.

Review: Definitions and properties

Remark: On certain domains the trigonometric functions are

invertible.

y

y = sin(x)

y

1

1

y = cos(x)

y y = tan(x)

- / 2

/2 x

0

-1 -1

y

y = csc(x)

y

/ 2

x

y = sec(x)

- / 2

/2 x

y y = cot(x)

1

- / 2

0

-1

/2 x

1

0

/ 2

-1

x

0

/ 2

x

Review: Definitions and properties

Remark: The graph of the inverse function is a reflection of the

original function graph about the y = x axis.

y y = arcsin(x)

/ 2

y y = arccos(x)

y y = arctan(x)

/ 2

-1 - / 2

1x

y y = arccsc(x)

/ 2

-1 0 1

x

- / 2

/ 2

-1

0

1x

y y = arcsec(x)

/ 2

-1 0 1

x

x

- / 2

y y = arccot(x)

/ 2

0

x

Review: Definitions and properties

Theorem

For all x [-1, 1] the following identities hold,

arccos(x) + arccos(-x) = ,

arccos(x) + arcsin(x) = .

2

Proof:

y

1

arccos(-x)

y

1

arcsin(x)

arccos(x)

-

-x = cos(-)

x = cos ()

x

x = sin(/2-) /2 -

x = cos ()

arccos(x)

x

Review: Definitions and properties

Theorem

For all x [-1, 1] the following identities hold,

arcsin(-x) = - arcsin(x), arctan(-x) = - arctan(x), arccsc(-x) = -arccsc(x).

Proof:

y y = arcsin(x)

/ 2

y y = arctan(x)

/ 2

y y = arccsc(x)

/ 2

-1 - / 2

1x

- / 2

x

-1 0 1

x

- / 2

Inverse trigonometric functions (Sect. 7.6)

Today: Derivatives and integrals.

Review: Definitions and properties. Derivatives. Integrals.

Derivatives of inverse trigonometric functions

Remark: Derivatives inverse functions can be computed with

f -1 (x) =

1 .

f f -1(x)

Theorem

1

The derivative of arcsin is given by arcsin (x) =

.

1 - x2

Proof: For x [-1, 1] holds

1

1

arcsin (x) =

=

sin arcsin(x) cos arcsin(x)

For x [-1, 1] we get arcsin(x) = y , , and the cosine is

22 positive in that interval, then cos(y ) = + 1 - sin2(y ), hence

1

1

arcsin (x) =

arcsin (x) =

.

1 - sin2 arcsin(x)

1 - x2

Derivatives of inverse trigonometric functions

Theorem

The derivative of inverse trigonometric functions are:

1

arcsin (x) =

,

1 - x2

1 arctan (x) = 1 + x2 ,

1

arcsec (x) =

,

|x| x2 - 1

1

arccos (x) = -

,

1 - x2

1 arccot (x) = - 1 + x2 ,

1

arccsc (x) = -

,

|x| x2 - 1

|x| 1, x R,

|x| 1.

Proof: arctan (x) =

tan

1 arctan(x )

,

cos2(y ) + sin2(y )

tan (y ) =

cos2(y )

tan (y ) = 1 + tan2(y ), y = arctan(x),

1 arctan (x) = 1 + x2 .

Derivatives of inverse trigonometric functions

Proof: arcsec (x) =

1 , for |x| 1.

sec arcsec(x)

Then y = arcsec(x) satisfies y [0, ] - {/2}. Recall,

1 sec (y ) =

cos(y )

sin(y ) = cos2(y ) ,

sin(y ) = + 1 - cos2(y ),

1 - cos2(y )

1

1 - cos2(y )

sec (y ) =

cos2(y )

= | cos(y )|

, | cos(y )|

1 sec (y ) =

| cos(y )|

1 cos2(y ) - 1 = | sec(y )|

sec2(y ) - 1.

1

We conclude: arcsec (x) =

.

|x| x2 - 1

Derivatives of inverse trigonometric functions

Example

Compute the derivative of y (x) = arcsec(3x + 7).

1

Solution: Recall the main formula: arcsec (u) =

.

|u| u2 - 1

3

Then, chain rule implies, y (x) =

.

|3x + 7| (3x + 7)2 - 1

Example

Compute the derivative of y (x) = arctan(4 ln(x)).

1 Solution: Recall the main formula: arctan (u) = 1 + u2 . Therefore, chain rule implies,

1

4

4

y (x) = 1 + 4 ln(x) 2 x

y

= x

1 + 16 ln2(x)

.

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