Derivatives of Exponential and Inverse Trig Functions Objective: To derive and use formulas for exponential and Inverse Trig Functions.

Derivatives of Exponential and Inverse Trig Functions

Objective: To derive and use formulas for exponential and Inverse

Trig Functions

Differentiability

• Geometrically, a function is differentiable at those points where its graph has a nonvertical tangent line. Since the graph of

is the reflection of the graph of about

the line y = x, it follows that the points where

is not differentiable are reflections of the points where the graph of f has a horizontal tangent line. Algebraically, will fail to be differentiable at a point (b, a) if

)(1 xfy

)(xfy 1f

1f

.0)(/ af

Differentiability

• We know that the equation of the tangent line to the graph of f at the point (a, b) is

))((/ axafby

Differentiability

• We know that the equation of the tangent line to the graph of f at the point (a, b) is

• To find the reflection on the line y = x we switch the x and the y, so it follows that the tangent line to the graph of at the point (b, a) is

))((/ axafby

))((/ ayafbx

1f

Differentiability

• We will rewrite this equation to make it

))((/ ayafbx )()(

1)(

/bx

afay

Differentiability

• We will rewrite this equation to make it

• This equation tells us that the slope

of the tangent line to the graph of at (b, a) is

or

If then

))((/ ayafbx )()(

1)(

/bx

afay

)()( /1 bf

1f

)(

1)()(

//1

afbf

))((

1)()(

1//1

bffbf

,)( baf abf )(1

Theorem 3.3.1

• The derivative of the inverse function of f is defined as:

))((

1)]([

1/1

xffxf

dx

d

Example

• Find the derivative of the inverse for the following function.

3

2)(

x

xf

Example


3

2)(

x

xf

32

)(

23

3

23

2

1

xxf

xy

yx

xy

Example


3

2)(

x

xf

32

)(

23

3

23

2

1

xxf

xy

yx

xy

21 2)]([

xxf

dx

d

Example


))((

1)]([

1/1

xffxf

dx

d

3

2)(

x

xf

Example


))((

1)]([

1/1

xffxf

dx

d

3

2)(

x

xf

22/

)3(

2

)3(

)1(2)0)(3()(

xx

xxf

Example


))((

1)]([

1/1

xffxf

dx

d

3

2)(

x

xf

22/

)3(

2

)3(

)1(2)0)(3()(

xx

xxf

2

)3(

)(

1 2

/

x

xf

Example


))((

1)]([

1/1

xffxf

dx

d

3

2)(

x

xf

22/

)3(

2

)3(

)1(2)0)(3()(

xx

xxf

2

)3(

)(

1 2

/

x

xf 2

22

1/

2

2

)33(

))((

1

xxffx

32

)(1

xxf

Theorem 3.3.1

• The derivative of the inverse function of f is defined as:

• We can state this another way if we let

then

))((

1)]([

1/1

xffxf

dx

d

)(1 xfy

dydxyfxf

dx

d

/

1

)(

1)]([

/1

Theorem 3.3.1


then )(1 xfy

dydxyfxf

dx

d

/

1

)(

1)]([

/1

23

42

42

2

3

3

y

yyx

xxy

dydx

23

1

/

12

ydydx

Theorem 3.3.1


then )(1 xfy

dydxyfxf

dx

d

/

1

)(

1)]([

/1

dxdy

dxdy

dxdy

y

y

yyx

xxy

23

1

231

42

42

2

2

3

3

23

1

/

12

ydydx

One-to-One

• We know that a graph is the graph of a function if it passes the vertical line test.

• We also know that the inverse is a function if the original function passes the horizontal line test.

One-to-One

• We know that a graph is the graph of a function if it passes the vertical line test.

• We also know that the inverse is a function if the original function passes the horizontal line test.

• If both of these conditions are satisfied, we say that the function is one-to-one. In other words, for every y there is only one x and for every x there is only one y.

Increasing/Decreasing

• If a function is always increasing or always decreasing, it will be a one-to-one function.

• Theorem 3.3.2 Suppose that the domain of a function f is an open interval I on which

or on which . Then f is one-to-one,

is differentiable at all values of x in the range of f and the derivative is given by

0)(/ xf0)(/ xf )(1 xf

))((

1)]([

1/1

xffxf

dx

d

Example 2

• Consider the function .

a)Show that f is one-to-one.

b)Show that is differentiable on the interval

c)Find a formula for the derivative of .

d)Compute .

1)( 5 xxxf

1f

),( 1f

)1()( /1f

Example 2



which is always positive, so the function is one-to-one.

1)( 5 xxxf

15)( 4/ xxf

Example 2




Since the range of f is , this is the domain of and from Theorem 3.3.2 is differentiable at all values of the range of f .

1)( 5 xxxf

1f

),( ),(

1f

Example 2





1)( 5 xxxf

1f

),( 1f

1)( 5 yyyfx

15 4 ydy

dx

Example 2





1)( 5 xxxf

1f

),( 1f

1)( 5 yyyfx

15 4 ydy

dx15

1

/

1)]([

41

ydydxxf

dx

d

Implicit

• Using implicit differentiation, we get

15 yyx

dx

dyy )15(1 4

dx

dy

y

15

14

Example 2




c)Find a formula for .

d)Compute

1)( 5 xxxf

1f

),( 1f

14

1

/1

15

1)1()(

xx ydx

dyf

)1()( /1f

Example 2





d)Compute• Since (0, 1) is a point on the function, the point

(1, 0) is on the inverse function.

1)( 5 xxxf

1f

),( 1f

14

1

/1

15

1)1()(

xx ydx

dyf

)1()( /1f

Example 2





d)Compute• Since (0, 1) is a point on the function, the point

(1, 0) is on the inverse function.

1)( 5 xxxf

1f

),( 1f

115

1)1()(

04

/1

yy

f

)1()( /1f 14

1

/1

15

1)1()(

xx ydx

dyf

Example 2A

• This is how we will use the other equation.• If f -1 is the inverse of f, write an equation of the

tangent line to the graph of y = f -1(x) at x = 6.

Example 2A


tangent line to the graph of y = f -1(x) at x = 6.• If f(1) = 6, f -1(6) = 1.

)6(1 xmy

Example 2A



)6(1 41 xy

4

1

)1(

1

))6((

1

))((

1)(

/1/1//1

fffxfff

You Try


tangent line to the graph of y = f -1(x) at x = 9.

Example 2A



)9(2 xmy

Example 2A



)9(2 21 xy

2

1

)2(

1

))9((

1

))((

1)(

/1/1//1

fffxfff

Derivatives of Exponential Functions

• We will use our knowledge of logs to find the derivative of . We are looking for dy/dx. xby


• We will use our knowledge of logs to find the derivative of . We are looking for dy/dx.

• We know that is the same as xby

xby

.log yx b



• We know that is the same as• We will take the derivative with respect to x and

simplify.

xby

xby

.log yx b

.log yx bdx

dy

by

ln

11

dx

dyby ln



• We know that is the same as• We will take the derivative with respect to x and

simplify.

• Remember that so

xby

xby

.log yx b

.log yx bdx

dy

by

ln

11

dx

dyby ln

xby bbdx

dy x ln


• This formula, works with any

base, so if the base is e, it becomes

but remember , so

bbbdx

d xx ln][

eeedx

d xx ln][

bbbdx

d xx ln][ xx ee

dx

d][

1ln e


• With the chain rule these formulas become:

dx

dubbb

dx

d uu ln][dx

duee

dx

d uu ][

Example 3

• Find the following derivatives:

]2[ x

dx

d

Example 3


2ln2]2[ xx

dx

d

Example 3


2ln2]2[ xx

dx

d

][ 2xedx

d

Example 3


2ln2]2[ xx

dx

d

xx eedx

d 22 2][

Example 3


2ln2]2[ xx

dx

d

xx eedx

d 22 2][

][3xe

dx

d

Example 3


2ln2]2[ xx

dx

d

xx eedx

d 22 2][

33 23][ xx exedx

d

Example 3


2ln2]2[ xx

dx

d

xx eedx

d 22 2][

33 23][ xx exedx

d

][ cos xedx

d

Example 3


2ln2]2[ xx

dx

d

xx eedx

d 22 2][

33 23][ xx exedx

d

xx xeedx

d coscos sin][

Example 4

• Use logarithmic differentiation to find ])1[( sin2 xxdx

d

Example 4

• Use logarithmic differentiation to find

• Let ])1[( sin2 xxy

])1[( sin2 xxdx

d

Example 4



])1[( sin2 xxdx

d

])1ln[(ln sin2 xxy

Example 4



])1[( sin2 xxdx

d

])1ln[(ln sin2 xxy

)]1ln[(sinln 2 xxy

Example 4


])1[( sin2 xxdx

d

)]1ln[(sinln 2 xxy

xxx

xx

dx

dy

ycos)]1ln[(

1

2sin

1 22

Example 4


])1[( sin2 xxdx

d

)]1ln[(sinln 2 xxy

xxx

xx

dx

dy

ycos)]1ln[(

1

2sin

1 22

])1][(cos)]1ln[(1

2[sin sin22

2xxxx

x

xx

dx

dy

Derivatives of Inverse Trig Functions

• We want to find the derivative of .sin 1 xy


• We want to find the derivative of

• We will rewrite this as and take the derivative.

.sin 1 xy

xy sin




.sin 1 xy

xy sin

1cos dx

dyy




.sin 1 xy

xy sin

1cos dx

dyy

ydx

dy

cos

1




.sin 1 xy

xy sin

1cos dx

dyy

ydx

dy

cos

1

)cos(sin

11 xdx

dy


• We need to simplify

• We will construct a triangle to help us do that. Remember that represents an angle where sin = x.

x1sin

)cos(sin

11 xdx

dy


• We need to simplify

• The cosine is the adjacent over the hypotenuse.

)cos(sin

11 xdx

dy

21

1

xdx

dy

Special Triangle

• Find )sin(cos 1 x

Special Triangle

• Find

• Again, we will construct a triangle where the cos = x to help solve this problem.

)sin(cos 1 x

21 1sin)sin(cos xx


dx

du

uu

dx

d

21

1

1][tan

dx

du

uu

dx

d

2

1

1

1][sin

dx

du

uu

dx

d

2

1

1

1][cos

dx

du

uuu

dx

d

1||

1][sec

2

1

dx

du

uu

dx

d

2

1

1

1][cot

dx

du

uuu

dx

d

1||

1][csc

2

1

Example 5

• Find dy/dx if:

)(sin 31 xy )(sec 1 xey

Example 5

• Find dy/dx if:


2

233

)(1

1x

xdx

dy

6

2

1

3

x

x

dx

dy

Example 5

• Find dy/dx if:


2

233

)(1

1x

xdx

dy

6

2

1

3

x

x

dx

dy

x

xxe

eedx

dy

1)(||

12

1

12

xedx

dy

Homework

• Section 3.3

• Pages 201-202

• 1, 5, 7, 9

• 15-27 odd

• 37-51 odd

Derivatives of Exponential and Inverse Trig Functions Objective: To derive and use formulas for exponential and Inverse Trig Functions.

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