Derivatives of Exponential and Logarithm Functions 10/17/2011 The - - PowerPoint PPT Presentation

derivatives of exponential and logarithm functions
SMART_READER_LITE
LIVE PREVIEW

Derivatives of Exponential and Logarithm Functions 10/17/2011 The - - PowerPoint PPT Presentation

Derivatives of Exponential and Logarithm Functions 10/17/2011 The Derivative of y = e x Recall! e x is the unique exponential function whose slope at x = 0 is 1: m=1 The Derivative of y = e x Recall! e x is the unique exponential function whose


slide-1
SLIDE 1

Derivatives of Exponential and Logarithm Functions

10/17/2011

slide-2
SLIDE 2

The Derivative of y = ex

Recall! ex is the unique exponential function whose slope at x = 0 is 1:

m=1

slide-3
SLIDE 3

The Derivative of y = ex

Recall! ex is the unique exponential function whose slope at x = 0 is 1:

m=1

lim

h→0

e0+h − e0 h = lim

h→0

eh − 1 h = 1

slide-4
SLIDE 4

The Derivative of y = ex...

lim

h→0

eh − 1 h = 1

slide-5
SLIDE 5

The Derivative of y = ex...

lim

h→0

eh − 1 h = 1 d dx ex = lim

h→0

ex+h − ex h

slide-6
SLIDE 6

The Derivative of y = ex...

lim

h→0

eh − 1 h = 1 d dx ex = lim

h→0

ex+h − ex h = lim

h→0

ex eh − 1

  • h
slide-7
SLIDE 7

The Derivative of y = ex...

lim

h→0

eh − 1 h = 1 d dx ex = lim

h→0

ex+h − ex h = lim

h→0

ex eh − 1

  • h

= ex lim

h→0

eh − 1 h

slide-8
SLIDE 8

The Derivative of y = ex...

lim

h→0

eh − 1 h = 1 d dx ex = lim

h→0

ex+h − ex h = lim

h→0

ex eh − 1

  • h

= ex lim

h→0

eh − 1 h = ex ∗ 1

slide-9
SLIDE 9

The Derivative of y = ex...

lim

h→0

eh − 1 h = 1 d dx ex = lim

h→0

ex+h − ex h = lim

h→0

ex eh − 1

  • h

= ex lim

h→0

eh − 1 h = ex ∗ 1 So d dx ex = ex

slide-10
SLIDE 10

The Chain Rule

Theorem

Let u be a function of x. Then d dx eu = eu du dx .

slide-11
SLIDE 11

Examples

Calculate... 1.

d dx e17x

2.

d dx esin x

3.

d dx e √ x2+x

slide-12
SLIDE 12

Examples

Calculate... 1.

d dx e17x = 17e17x

2.

d dx esin x = cos(x)esin x

3.

d dx e √ x2+x = 2x+1 2 √ x2+x e √ x2+1

slide-13
SLIDE 13

Examples

Calculate... 1.

d dx e17x = 17e17x

2.

d dx esin x = cos(x)esin x

3.

d dx e √ x2+x = 2x+1 2 √ x2+x e √ x2+1

Notice, every time: d dx ef (x) = f ′(x)ef (x)

slide-14
SLIDE 14

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!

slide-15
SLIDE 15

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x = ⇒ ey = x

slide-16
SLIDE 16

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x = ⇒ ey = x Take a derivative of both sides of ey = x to get dy dx ey = 1

slide-17
SLIDE 17

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x = ⇒ ey = x Take a derivative of both sides of ey = x to get dy dx ey = 1 So dy dx = 1 ey

slide-18
SLIDE 18

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x = ⇒ ey = x Take a derivative of both sides of ey = x to get dy dx ey = 1 So dy dx = 1 ey = 1 eln(x)

slide-19
SLIDE 19

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x = ⇒ ey = x Take a derivative of both sides of ey = x to get dy dx ey = 1 So dy dx = 1 ey = 1 eln(x) = 1 x

slide-20
SLIDE 20

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x = ⇒ ey = x Take a derivative of both sides of ey = x to get dy dx ey = 1 So dy dx = 1 ey = 1 eln(x) = 1 x

d dx ln(x) = 1 x

slide-21
SLIDE 21

Does it make sense?

d dx ln(x) = 1 x

f (x) = ln(x)

1 2 3 4

  • 1

1

f (x) = 1

x

1 2 3 4 1 2 3

slide-22
SLIDE 22

Examples

Calculate 1.

d dx ln x2

2.

d dx ln(sin(x2))

slide-23
SLIDE 23

Examples

Calculate 1.

d dx ln x2 = 2x

x2 = 2 x 2.

d dx ln(sin(x2)) = 2x cos(x2)

sin(x2)

slide-24
SLIDE 24

Examples

Calculate 1.

d dx ln x2 = 2x

x2 = 2 x 2.

d dx ln(sin(x2)) = 2x cos(x2)

sin(x2) Notice, every time: d dx ln(f (x)) = f ′(x) f (x)

slide-25
SLIDE 25

The Calculus Standards: ex and ln x

To get the other derivatives: ax = ex ln a loga x = ln x ln a

slide-26
SLIDE 26

The Calculus Standards: ex and ln x

To get the other derivatives: ax = ex ln a loga x = ln x ln a For example: d dx 2x

slide-27
SLIDE 27

The Calculus Standards: ex and ln x

To get the other derivatives: ax = ex ln a loga x = ln x ln a For example: d dx 2x = d dx ex ln(2)

slide-28
SLIDE 28

The Calculus Standards: ex and ln x

To get the other derivatives: ax = ex ln a loga x = ln x ln a For example: d dx 2x = d dx ex ln(2) = ln(2) ∗ ex ln(2)

(ln(2) is a constant!!!)

slide-29
SLIDE 29

The Calculus Standards: ex and ln x

To get the other derivatives: ax = ex ln a loga x = ln x ln a For example: d dx 2x = d dx ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

slide-30
SLIDE 30

The Calculus Standards: ex and ln x

To get the other derivatives: ax = ex ln a loga x = ln x ln a For example: d dx 2x = d dx ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try: d dx log2(x)

slide-31
SLIDE 31

The Calculus Standards: ex and ln x

To get the other derivatives: ax = ex ln a loga x = ln x ln a For example: d dx 2x = d dx ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try: d dx log2(x) = 1 ln(2) ∗ x

slide-32
SLIDE 32

Differential equations

Suppose y is some mystery function of x and satisfies the equation y′ = ky Goal: What is y??

slide-33
SLIDE 33

Differential equations

Suppose y is some mystery function of x and satisfies the equation y′ = ky Goal: What is y??

  • 1. If k = 1, then y = ex has this property and thus solves the

equation.

slide-34
SLIDE 34

Differential equations

Suppose y is some mystery function of x and satisfies the equation y′ = ky Goal: What is y??

  • 1. If k = 1, then y = ex has this property and thus solves the

equation.

  • 2. For any k, y = ekx solves the equation too!
slide-35
SLIDE 35

Differential equations

Suppose y is some mystery function of x and satisfies the equation y′ = ky Goal: What is y??

  • 1. If k = 1, then y = ex has this property and thus solves the

equation.

  • 2. For any k, y = ekx solves the equation too!

This equation,

d dx y = ky is an example of a differential equation.