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Lecture #12 – Thursday, October 16, 2003

DIFFERENTIATION

• What is the slope of the secant line PQ?

(

)

(

)

( )

(

)

(

)

h

cfhcf

chc

cfhcf−+

=

−+

−+ .

• Idea: Get the slope of the tangent line as a limit of slopes of secant lines.

• The slope of the tangent line at

c

x

=

“ought” to be

(

)

(

)

h

cfhcf

h

−+

→0

lim.

Example

(

)

xxf= at 0

=

x

• This DOESN’T have a well defined tangent line

Definition

• f is differentiable at

c

x

=

if the limit

(

)

(

)

h

cfhcf

h

−+

→0

lim exists. If it does, we call it the derivative of f at

c and we denote it by

(

)

cf

′

.

Geometrically

•

(

)

cf

′

is the slope of the tangent line going through

(

)

(

)

cfc,.

• What is the equation for tangent line?

(

)

(

)

(

)

cxcfcfy−

′

=−

P

c

Q

f

(

)

(

)

hcfhc++ ,

(

)

(

)

cfc,

c + h

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MAT137Y1b.doc

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Lecture #13 – Tuesday, October 21, 2003

Example

• For function

(

)

2

xxf=, the derivative of f at 2

=

c is

( )

(

)

(

)

( )

44lim

22

lim2

0

22

0

=+=

−+

=

′

→→ h

h

h

f

hh

.

• The derivative of f is itself a function – for

(

)

2

xxf= repeat the same calculation for any value of c.

•

( )

(

)

(

)

(

)

(

)

( )

chc

h

chchc

h

chc

h

cfhcf

cf

hhhh

22lim

2

limlimlim

0

222

0

22

00

=+=

−++

=

−+

=

−+

=

′

→→→→

• At any fixed value of x,

(

)

xxf2=

′

.

Definition

• The derivative of f is a function, denoted f

′

, and

( )

(

)

(

)

h

xfhxf

xf

h

−+

=

′

→0

lim, if it exists.

Terminology

• To differentiate a function is to find the derivative.

• Notice: The function f has to be defined in the interval

(

)

δ+δ−xx , in order for

(

)

xf

′

to be defined.

Example

• Actually, even if f is continuous on

(

)

δ+δ−xx ,, it doesn’t mean

(

)

xf

′

is defined.

• Consider

(

)

0,== cxxf

Theorem

• If f is differentiable at x, the f is continuous.

• “Being differentiable is ‘better’ than being continuous.”

• Proof:

• Because f is differentiable at x,

(

)

(

)

( )

xf

h

xfhxf

h

′

=

−+

→0

lim

•

( ) ( )( )

(

)

(

)

( )

00limlim

00

=⋅

′

=⋅

−+

=−+ →→ xfh

h

xfhxf

xfhxf

hh

• So f is continuous.

DIFFERENTIATION RULES

Building Blocks

• If

(

)

cxf= (a constant function), then

(

)

0=

′

xf for all x.

• If

(

)

xxf=, then

(

)

1=

′

xf for all x.

f(x)

x

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Page 3 of 38

Theorem: Sums and Scalar Multiples

• Let f, g be differentiable at x and α a constant.

• Then

(

)

gf+ and fα are differentiable, then

•

( ) ( ) ( ) ( )

xgxfxgf′

+

′

=

′

+

•

( ) ( ) ( )

xfxf′

α=

′

α

• Proof:

•

( ) ( )

(

)

(

)

(

)

(

)

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

xgxf

h

xghxg

h

xfhxf

h

xgxfhxghxf

h

xgfhxgf

xgf

h

h

h

′

+

′

=

−+

+

−+

=

−−+++

=

+

−

+

+

=

′

+

→

→

→

0

0

0

lim

lim

lim

Example

• If

(

)

xxf10=,

(

)

10=

′

xf

Theorem: Differences and Linear Combinations

• From the Sums and Scalar Multiples rule,

•

( ) ( ) ( ) ( )

xgxfxgf′

−

′

=

′

−

•

( ) ( ) ( ) ( ) ( )

xfxfxfxfff nnnn

′

⋅α++

′

⋅α+

′

⋅α=

′

α++α+α...... 22112211

Theorem: Product Rule

• If f and g are differentiable at x, then gf⋅ is differentiable and

( ) ( ) ( ) ( ) ( ) ( )

xgxfxgxfxgf′

⋅+⋅

′

=

′

⋅

• Proof:

•

( ) ( )

(

)

(

)

(

)

(

)

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

xgxfxgxf

gf

h

xghxg

xfhxg

h

xfhxf

h

xgxfhxgxfhxgxfhxghxf

h

xgxfhxghxf

h

xgfhxgf

xgf

hh

h

h

h

′

+

′

=

−+

⋅++⋅

−+

=

⋅−+⋅++⋅−+⋅+

=

⋅−+⋅+

=

⋅−+⋅

=

′

⋅

→→

→

→

→

.continuous are they able,differenti are , Because

limlim

lim

lim

lim

00

0

0

0

Theorem: Power Rule

• Using the Product Rule, we derive the Power Rule.

• For Z

∈

>

nn ,0, if

(

)

n

xxf=, then

(

)

1−

=

′n

nxxf.

• Proof (by induction):

• True for 1=k

• Assume

(

)

1−

=

′

kk kxx . Prove for 1+k:

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