Kalkulus 03-Derivat

8/2/2019 Kalkulus 03-Derivat

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0

limh

f a h f a

h

is called the derivative of at .f a

We write:

0limh

f a h f af x

h

The derivative off with respect toxis

There are many ways to write the derivative of y f x

3.1 Derivative of a Function


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f x f prime x or the derivative of f with respectto x

y

y primedy

dxdee why dee ecks or the derivative of y with

respect to x

df

dx dee eff dee ecks or the derivative of f withrespect to x

d f xdx dee dee ecks uv eff uv ecks or the derivative

of f of x( of of )d dx f x



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dx does not mean dtimesx !

dy does not mean dtimesy !



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dy

dxdoes not mean !dy dx

(except when it is convenient to think of it as division.)

df

dxdoes not mean !df dx

(except when it is convenient to think of it as division.)



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(except when it is convenient to treat it that way.)

d

f xdx does not mean times !

d

dx f x



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The derivative is

the slope of theoriginal function.

The derivative is definedat the end points of afunction on a closedinterval.

0

1

2

3

4

1 2 3 4 5 6 7 8 9

y f x

-2

-1

0

1

2

3

1 2 3 4 5 6 7 8 9

y f x



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-3

-2

-1

0

1

2

3

4

5

6

- 3 - 2 - 1 1 2 3x

2 3y x

2 2

0

3 3limh

x h xy

h

2y x

-6

-5-4

-3

-2

-10

1

23

4

5

6

-3 -2 -1 1 2 3x

0lim2h

y x h



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A function is differentiable if it has aderivative everywhere in its domain. Itmust be continuous and smooth.Functions on closed intervals must haveone-sided derivatives defined at the endpoints.



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To be differentiable, a function must be continuousand smooth.Derivatives will fail to exist at:

corner

f x x

cusp

2

3 f x x

vertical tangent

3 f x x

discontinuity

1, 0

1, 0

x f x

x

3.2 Differentiability


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Most of the functions we study in calculus will be differentiable.



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There are two theorems on page 110:

If f has a derivative atx = a, thenf is continuous atx = a.

Since a function must be continuous to have a derivative,if it has a derivative then it is continuous.



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1

2f a

3f b

Intermediate Value Theorem for Derivatives

Between a and b, must take

on every value between and .

f1

23

If a and b are any two points in an interval on which f isdifferentiable, then takes on every value between

and .

f f a

f b



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If the derivative of a function is its slope, then for aconstant function, the derivative must be zero.

0d

cdx

example: 3y

0y

The derivative of a constant is zero.

3.3 Rules for Differentiation


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We saw that if , .2

y x 2y x

This is part of a pattern.

1n nd

x nxdx

examples:

4 f x x

34 f x x

8y x

78y x

power rule



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1n nd

x nx

dx

3.3 Rules for DifferentiationProof:

h

xhx

xdx

dnn

h

n

)(

lim0

h

xhhnxxx

dx

dnnnn

h

n

...lim

1

0

h

hhnxx

dx

d nn

h

n

...lim

1

0

1

0

lim

n

h

n nxx

dx

d


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d du

cu cdx dx

examples:

1n nd cx cnxdx

constant multiple rule:

5 4 47 7 5 35d

x x xdx



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(Each term is treated separately)

d du

cu cdx dx

constant multiple rule:

sum and difference rules:

d du dv

u vdx dx dx

d du dv

u vdx dx dx

4 12 y x x

3

4 12y x

4 22 2 y x x

3

4 4

dy

x xdx



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Find the horizontal tangents of:4 22 2 y x x 34 4

dyx x

dx

Horizontal tangents occur when slope = zero.34 4 0x x

3 0x x

2 1 0x x 1 1 0 x x x

0, 1, 1x

Substituting the x values into theoriginal equation, we get:

2, 1, 1 y y y (The function is even, so weonly get two horizontaltangents.)



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-2

-1

0

1

2

3

4

-2 -1 1 2

4 22 2 y x x

2y

1y



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-2

-1

0

1

2

3

4

-2 -1 1 2

4 22 2 y x x

First derivative(slope) is zero at:

0, 1, 1x

34 4

dyx x

dx



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product rule:

d dv duuv u vdx dx dx

Notice that this is not just theproduct of two derivatives.

This is sometimes memorized as: d uv u dv v du 2 33 2 5

d x x x

dx

5 3 3

2 5 6 15

d

x x x xdx

5 32 11 15d

x x xdx

4 2

10 33 15x x

2 3x 26 5x 32 5x x 2x

4 2 2 4 26 5 18 15 4 10 x x x x x

4 2

10 33 15x x



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product rule:

d dv duuv u vdx dx dx


Proof

h

xvxuhxvhxuuv

dx

d

h

)()()()(lim)(

0

add and subtract u(x+h)v(x)in the denominator

h

xvhxuxvhxuxvxuhxvhxuuv

dx

d

h

)()()()()()()()(lim)(

0

h

xuhxuxvxvhxvhxuuv

dx

d

h

)()()()()()(lim)(

0

dx

duv

dx

dvuuv

dx

d)(


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quotient rule:

2

du dvv u

d u dx dx

dx v v

or 2u v du u dv

dv v

3

2

2 5

3

d x x

dx x

2 2 3

22

3 6 5 2 5 2

3

x x x x x

x



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Higher Order Derivatives:dy

y

dx

is the first derivative of y with respect to x.

2

2

dy d dy d yy

dx dx dx dx

is the second derivative.

(y double prime)

dy

y dx

is the third derivative.

4 dy ydx

is the fourth derivative.

We will learnlater what thesehigher orderderivatives are

used for.



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3.3 Rules for DifferentiationSuppose u and v are functions that are differentiable at

x = 3, and that u(3) = 5, u(3) = -7, v(3) = 1, and v(3)= 4.

Find the following atx = 3 :

)(.1 uvdx

d'')( vuuvuv

dx

d 8)7)(1()3(5

vu

dxd.2

2''

vuvvu

vu

dxd

21

)4)(5()7)(1( 27

u

v

dx

d.3

2

''

u

vuuv

u

v

dx

d

2

5

)7)(1()4)(5(

25

27


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hi

ho

dx

d

))((

)()()()(

hoho

hidhohodhi


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Consider a graph of displacement (distance traveled) vs. time.

time (hours)

distance(miles)

Average velocity can be found bytaking:

change in positionchange in time

s

t

t

sA

B

ave

f t t f t sV

t t

The speedometer in your car does not measure averagevelocity, but instantaneous velocity.

0

limt

f t t f t dsV t

dt t

(The velocity at one

moment in time.)

3.4 Velocity and other Rates

of Change


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of Change

Velocity is the first derivative of position.

Acceleration is the second derivative

of position.


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Example: Free Fall Equation

21 2

s g t

GravitationalConstants:

2ft32

secg

2

m9.8

sec

g

2

cm980

secg

21 322

s t

216s t 32ds

V tdt

Speed is the absolute value of velocity.


of Change


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Acceleration is the derivative of velocity.

dv

a dt

2

2

d s

dt example:32v t

32a If distance is in: feet

Velocity would be in: feetsec

Acceleration would be in:

ft

sec

sec

2

ft

sec


of Change


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time

distance

acc posvel pos &increasing

acc zerovel pos &constant

acc negvel pos &decreasing

velocityzero

acc negvel neg &decreasing acc zero

vel neg &constant

acc posvel neg &increasing

acc zero,velocity zero


of Change


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Rates of Change:

Average rate of change = f x h f xh

Instantaneous rate of change =

0limh

f x h f xf x

h

These definitions are true for any function.

( x does not have to represent time. )


of Change


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For a circle: 2A r

2dA d rdr dr

2dA

rdr

Instantaneous rate of change of the area withrespect to the radius.

For tree ring growth, if the change in area is constant then dr

must get smaller asr

gets larger.

2dA r dr


of Change


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from Economics:

Marginal cost is the first derivative of the cost function, andrepresents an approximation of the cost of producing onemore unit.


of Change


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Example 13:Suppose it costs: 3 26 15c x x x x to producex stoves. 23 12 15c x x x If you are currently producing 10 stoves,the 11th stove will cost approximately:

210 3 10 12 10 15c 300 120 15

$195

marginal cost

The actual cost is: 11 10C C

3 2 3 211 6 11 15 11 10 6 10 15 10

770 550 $220 actual cost


of Change


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Note that this is not agreat approximationDont let that bother you.

Marginal cost is a linear approximation of a curvedfunction. For large values it gives a good approximationof the cost of producing the next item.


of Change


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of Change


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2

0

2

Consider the function siny We could make a graph of the slope:

slope

1

0

1

0

1

Now we connect the dots!The resulting curve is a cosine curve.

sin cosd

x x

dx

3.5 Derivatives of

Trigonometric Functions


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3.5 Derivatives of


hxhxx

dxd

hsin)sin(limsin

0

h

xxhhxx

dx

d

h

sincossincossinlimsin

0

h

xh

h

hx

xdx

d

hh

cossin

lim

)1(cossin

limsin 00

h

xhhxx

dx

d

h

cossin)1(cossinlimsin

0

Proof


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3.5 Derivatives of


h

xh

h

hxx

dx

d

hh

cossinlim

)1(cossinlimsin

00

= 0 = 1

sin cosd

x xdx


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3.5 Derivatives of


hxhxx

dxd

hcos)cos(limcos

0

h

xxhhxx

dx

d

h

cossinsincoscoslimcos

0

h

xh

h

hx

xdx

d

hh

sinsin

lim

)1(coscos

limcos 00

h

xhhxx

dx

d

h

sinsin)1(coscoslimcos

0

Find the derivative of cos x


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3.5 Derivatives of

Trigonometric Functions= 0 = 1

h

xh

h

hxx

dx

d

hh

sinsinlim

)1(coscoslimcos

00

cos sin

d

x xdx


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We can find the derivative of tangentx by using thequotient rule.

tand xdx

sin

cos

d x

dx x

2

cos cos sin sin

cos

x x x x

x

2 2

2cos sin

cosx x

x

2

1

cos

x2

sec x

2tan secd x xdx

3.5 Derivatives of



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Derivatives of the remaining trig functionscan be determined the same way.

sin cosd x xdx

cos sind

x xdx

2tan sec

dx x

dx

2cot cscd x x

dx

sec sec tand

x x xdx

csc csc cotd

x x xdx

3.5 Derivatives of



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3.5 Derivatives of


Jerk A sudden change in acceleration

Definition JerkJerk is the derivative of acceleration. If a bodys positionat time tis s(t), the bodys jerk at time t is

3

3

2

2

)(dt

sd

dt

vd

dt

datj


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3.5 Derivatives of



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Consider a simple composite function:6 10y x

2 3 5y x If 3 5u x

then 2y u

6 10y x 2y u 3 5u x

6dy

dx 2

dy

du 3

du

dx

dy dy du

dx du dx

6 2 3

3.6 Chain Rule


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dy dy du

dx du dx Chain Rule:

example: sin f x x 2 4g x x Find: at 2 f g x

cos f x x 2g x x 2 4 4 0g

0 2f g cos 0 2 2 1 4 4

3.6 Chain Rule

If is the composite of and ,then:

f g y f u u g x

atat xu g x f g f g )('))((' xgxgf


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2sin 4 f g x x

2sin 4y x siny u 2 4u x

cosdy

udu

2du

xdx

dy dy du

dx du dx

cos 2

dy

u xdx

2cos 4 2dy

x xdx

2cos 2 4 2 2dy

dx

cos 0 4dydx

4dy

dx

3.6 Chain Rule


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Here is a faster way to find the derivative:

2sin 4y x

2 2cos 4 4d

y x xdx

2cos 4 2 y x x

Differentiate the outside function...

then the inside function

At 2, 4x y

3.6 Chain Rule


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2cos 3d

xdx

2

cos 3d

xdx

2 cos 3 cos 3dx xdx

2cos 3 sin 3 3d

x x x

dx

2cos 3 sin 3 3x x

6cos 3 sin 3x x

The chain rule can be usedmore than once.

(Thats what makes the

chain in the chain rule!)

3.6 Chain Rule


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Derivative formulas include the chain rule!

1n n

d duu nu

dx dx sin cos

d duu u

dx dx

cos sind du

u udx dx

2tan secd du

u udx dx

etcetera

3.6 Chain Rule


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3.6 Chain RuleFind

)3cos( 2 xxy )16)(3sin( 2 xxxdx

dy

))sin(cos(xy

)24(cos 33 xxy

)sin)(cos(cos xx

dx

dy

)212))(24sin()(24(cos3 2332 xxxxxdx

dy

))24sin()(24(cos)636(3322

xxxxxdx

dy

dx

dy


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The chain rule enables us to find the slope ofparametrically defined curves:

dy dy dx

dt dx dt

dydydt

dx dx

dt

The slope of a parametrizedcurve is given by:

dy

dy dtdxdx

dt

3.6 Chain Rule


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These are the equations for

an ellipse.

Example:

3cosx t 2siny t

3sindx

tdt

2cosdy

tdt

2cos

3sin

dy t

dx t

2

cot3

t

3.6 Chain Rule


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2 2 1x y This is not a function,but it would still benice to be able to find

the slope.2 2 1d d d

x ydx dx dx

Do the same thing to both sides.

2 2 0

dy

x y dx

Note use of chain rule.

2 2dy

y xdx

22

dy xdx y

dy xdx y

3.7 Implicit Differentiation


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22 sin y x y

22 sind d d

y x ydx dx dx

This cant be solved fory.

2 2 cosdy dy

x ydx dx

2 cos 2dy dyy xdx dx

22 cosdy

xy

dx

2

2 cos

dy x

dx y

This technique is calledimplicit differentiation.

1 Differentiate both sides w.r.t.x.

2 Solve for .dy

dx



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Implicit Differentiation Process

1. Differentiate both sides of the equation with respect tox.2. Collect the terms with dy/dx on one side of the equation.3. Factor out dy/dx .4. Solve for dy/dx .


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Find the equations of the lines tangent and normal to the

curve at .2 2 7 x xy y ( 1,2)

2 2 7 x xy y

2 2 0

dydy

x yx y dxdx

Note product rule.

2 2 0dy dy

x x y ydx dx

22dy

y xy x dx

2

2

dy y x

dx y x

2 2 1

2 2 1m

2 2

4 1

4

5



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Find the equations of the lines tangent and normal to the

curve at .2 2 7 x xy y ( 1,2)

4

5m

tangent:

4

2 15

y x

4 42 5 5y x

4 14

5 5

y x

normal:

5

2 14

y x

5 52 4 4y x

5 3

4 4y x



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Find if .

2

2

d y

dx3 22 3 7x y

3 2

2 3 7x y 26 6 0 x y y

26 6 y y x

26

6

xy

y

2xy

y

2

2

2y x x yy

y

2

2

2x xy y

y y

2 2

2

2x xyy

x

yy

4

3

2x xy

y y

Substituteback into theequation.

y



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Rational Powers of Differentiable Functions

Power Rule for Rational Powers ofx

If n is any rational number, then

1 nn nxxdxd


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Proof: Letp and q be integers with q > 0.

q

p

xy pq

xy

Raise both sides to the q power

Differentiate with respect tox

11 pq pxdx

dyqy Solve for dy/dx


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1

1

q

p

qy

px

dx

dy Substitute for y

1/

1

)(

qqp

p

xq

px

dx

dyRemove parenthesis

qpp

p

qx

px

dx

dy/

1

Subtract exponents

q

px

dx

dyqppp )/(1

1)/( qpx

q

p

dx

dy

3 8 D i ti f I


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Becausex andy are

reversed to find the

reciprocal function, the

following pattern always

holds:86420

8

6

4

2

0

x

y

x

y

2y x

y x

4m 2,4

4,21

4m

Slopes are

reciprocals.

Derivative Formula for Inverses:

df

dx df

dx x f a

x a

1 1

( )

evaluated at ( )f a

is equal to the reciprocal of

the derivative of ( )f x

evaluated at .a

The derivative of 1( )f x

3.8 Derivatives of InverseTrigonometric Functions

3 8 D i ti f I


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-1.5

-1

-0.5

0

0.5

1

1.5

-1.5 -1 -0.5 0.5 1 1.5

siny x

1siny xWe can use implicit

differentiation to find:1sin

dx

dx

1siny x

siny x

sind d

y xdx dx

cos 1dy

ydx

1

cos

dy

dx y


3 8 D i ti f I


69/86

We can use implicitdifferentiation to find:

1sind

xdx

1siny x

sin y x sind dy xdx dx

cos 1dy

y

dx

1

cos

dy

dx y

2 2sin cos 1y y

2 2

cos 1 siny y 2cos 1 siny y

But

2 2

y

so is positive.cos y

2cos 1 siny y

2

1

1 sin

dy

dx y

2

1

1

dy

dx x


3 8 D i ti f I


70/86

1siny x

1

cos

dy

dx y


)cos(sin

11xdx

dy

x

1

sin

x

1

21 x21

1

xdx

dy

xy sin

1cos dxdyy

3 8 D i ti f I


71/86


)(tansec

112xdx

dy

x

1

tan

x

1

21 x

211xdx

dy

xy tan

1sec2 dxdyy

Find xdx

d 1tan

xy 1tan

ydx

dy2

sec

1

3 8 D i ti f I


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)tan(sec)sec(sec1

11xxdx

dy

x1sec

x

1

12 x

1||

1

2

xxdx

dyxy sec

1tansec dxdyyy

Find xdx

d 1sec

xy 1sec

yydx

dy

tansec

1

3 8 D i ti f I


73/86

1

2

1sin

1

d duu

dx dxu

121tan 1

d duudx u dx

1

2

1sec

1

d duu

dx dxu u

1

2

1cos

1

d duu

dx dxu

1

21cot

1d duu

dx u dx

1

2

1csc

1

d duu

dx dxu u

1 1cos sin2

x x 1 1cot tan

2x x

1 1csc sec2

x x


3 8 De i ti e of In e e


74/86

Your calculator contains allsix inverse trig functions.However it is occasionally

still useful to know thefollowing:

1 1 1sec cosxx

1 1cot tan2

x x

1 1 1csc sinxx


3 8 Derivatives of Inverse


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Find

)3(cos 21 xy

422 916)6(

)3(1(1

xxx

xdxdy

x

y1

cot1

xxy 1sec

1

11

11

122

2

xx

xdx

dy

)1)((sec1||

1 12

xxx

xdx

dy

dx

dy

3 9 D i ti f E ti l d


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-1

0

1

2

3

-3 -2 -1 1 2 3x

Look at the graph ofx

y e

The slope atx = 0appears to be 1.

If we assume this to betrue, then:

0 0

0lim 1

h

h

e e

h

definition of derivative

3.9 Derivatives of Exponential and

Logarithmic Functions

3 9 D i ti f E ti l


77/86

Now we attempt to find a general formula for the

derivative of using the definition.xy e

0

lim x h x

x

h

d e ee

dx h

0lim

x h x

h

e e e

h

0

1lim

hx

h

ee

h

0

1lim

hx

h

ee

h

1x

e

xe

This is the slope atx = 0, which

we have assumed to be 1.

3.9 Derivatives of Exponential

and Logarithmic Functions


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x xd

e edx

3 9 Derivatives of Exponential


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xe is its own derivative!

If we incorporate the chain rule: u ud due e

dx dx

We can now use this formula to find the derivative ofxa

3.9 Derivatives of Exponentialand Logarithmic Functions



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xd

adx

lnx

ad edx

lnx ad edx

ln lnx ad

e x adx

Incorporating the chain rule:

lnu ud du

a a adx dx


aaa

dx

d xx ln



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So far today we have:

u ud due e

dx dx lnu u

d dua a a

dx dx

Now it is relatively easy to find the derivative of .ln x




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lny xye x

yd d

e x

dx dx

1ydy

e

dx

1y

dy

dx e

1ln

dx

dx x

1ln

d duu

dx u dx




83/86

To find the derivative of a common log function, youcould just use the change of base rule for logs:

logd xdx

lnln10

d xdx

1 lnln10

d xdx

1 1ln10 x

The formula for the derivative of a log of any base

other than e is:

1log

ln

a

d duu

dx u a dx




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u ud due e

dx dx

lnu ud du

a a adx dx

1logln

ad duudx u a dx

1lnd duudx u dx




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x

ey

2

2

3xy

3

ln xy

)(sin 41 xey

Findy

x

ey

2

2')2)(3ln(3'

2

xy x

xxxy3

)3(1

'2

3

)4)((

)(1

1' 4

24

x

xe

e

y



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Logarithmic differentiation

Used when the variable is in the base and the exponent

y = xx

ln y = ln xx

ln y = x ln x

xxd

dyln

11

xydx

dyln1

xxdxdy x ln1

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Kalkulus 03-Derivat