calculus i chapter three1. 2 calculus timeline: descartes 1596-1650 cavalieri 1598-1647 fermat...

Calculus I Chapter Three 1


Calculus Timeline: Descartes 1596-1650Cavalieri 1598-1647Fermat 1601-1665Wallis 1616-1703Barrow 1630-1677Gregory 1638-1675Newton 1642-1727Leibniz 1646-1716


The Tangent Problem:

were called infinitesimals, very small numbers, but not zero!

This idea was problematic because infinitesimals were called ghosts of

departed quantities by Berkley!

dx and dy

GeoGebra

Slope of the tangent line:

dyslope

dx


0


: What is the slope of the tangent line at ?

dyslope

dxExample x

3

3 3 2 2 30 0 0 0

3 2 2 30 0 0 0 0

2 3

3 20 0 0 0

3 2 30 0 0 0 0 0 0

( ) 2 .

: ( ) 3 3

( ) 3 3 2 2

, :

( ) 3 2 2

( ) ( ) 3 2 2 2

Let y f x x x

Note x dx x x dx x dx dx

f x dx x x dx x dx dx x dx

Since dx is very small dx and dx go to zero

f x dx x x dx x dx

dy f x dx f x x x dx x dx x x

dy

20

2 20 0 0 0

3 2

(3 2 ) 3 2

x dx dx

dydy x x dx x x

dx

5Calculus I Chapter Three

Explain how slopes of secant lines approach the slopes of the tangent line at a point.(Definition(1))

Definition(1) Derivative at a point GeoGebra file

Slope of the secant line:

( ) ( )f x f a

x a


( ) ( )limx a

f x f a

x a


Explain how slopes of secant lines approach the slopes of the tangent line at a point. (Definition(2))

Slope of the secant line:

( ) ( )f x h f x

h

0


( ) ( )limh

f x h f x

h

Definition(2) Derivative at a point GeoGebra file


The Derivative Function


3( )f x xAverage Rates

The pattern of the average rates looks quadratic!

2( ) 3f x x

1 2

2 3

3 44 55 6

6 7

(8 1)7

(2 1)

x Average Rate

1 7

2 19

3 37

4 61

5 91

6 127

3 2

2

The derivative of

( ) is 3 .

( ) 3 is a function called

the derivative function.

f x x x

f x x


The derivative can be interpreted in two ways:• the slope of the tangent line to the graph of the relation at the given x value.• the rate of change of the function with respect to x at the given x value.



( ) ( )limx a

f x f a

x a


0


( ) ( )limh

f x h f x

h


a-D b-C c-B d-A


Differentiable Implies Continuous

( ) is not defined at f x a( ) is not defined at f x a

( ) is not defined at f x a

( ) is not defined at f x a


0 0 0

( )

( ) ( )( ) lim lim lim 0 0

h h h

f x c

f x h f x c cf x

h h


1 11

0

What is the derivative of the sum of differentiable functions?

Let ( ) ( ) ( ), does ( ) exist?

( ) ( )( ) lim

x

h x f x g x h x

h x x h xh x

x

1 1 1

1 1 1

1 1

1 1 1 1

1 1 1 1

1 1 1 1

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ( ) ( ))

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

h x x f x x g x x

h x f x g x

h x x h x

xf x x g x x f x g x

xf x x f x g x x g x


x x


1 11 0

1 1 1 1

0

1 1 1 1

0 0

1 1

( ) ( )( ) lim

( ) ( ) ( ) ( )lim

( ) ( ) ( ) ( )lim lim

( ) ( )

x

x

x x

h x x h xh x


x xf x x f x g x x g x

x xf x g x

What is the derivative of the sum of differentiable functions?

Let ( ) ( ) ( ), does ( ) exist? Yes!h x f x g x h x


2

Is the derivative of a product the product of the derivatives?

( ( ) ( )) ( ) ( )

( ) 1 ( ) 1

( ) 1 ( ) 1

( ) ( ) 1

But:

( ) ( ) 1

( ( ) ( )) 2

Is the deriv

f x g x f x g x

Example

f x x g x x

f x g x

f x g x

f x g x x

f x g x x

ative of a product the product

of the derivative? !NO


1 11

0

What is the derivative of the product of

differentiable functions?

Let ( ) ( ) ( ), does ( ) exist?

We use the definition of derivative:

( ) ( )( ) lim

Using the definition we'll p

x

h x f x g x h x

h x x h xh x

x

1 1 1 1 1

rove that:

If ( ) ( ) ( ), then

( ) ( ) ( ) ( ) ( )

h x f x g x

h x g x f x g x f x


1 1 1

1 1 1

1 1 1

1 1 1 1

1 1 1

1 1 1 1

1 1 1 11 1

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) (

( ) ( )

) (

( )

) ( ( ) ( ))

( ) ( ) ( ( ) ( ))

(

( )

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

h x x f x x g x x

h x f x g x

h x x h x f x x g x x f x g xx x

f x x g x x f x g xx

g x x f x

f x g x x f x g x x

f x f x g xx xx

x g

1 11

1

1 11

1 110

1 1 1 1

11

1 1

11

1

1

11

(( ( ) ( ))( )

(( ( ) ( ))lim ( )

( ) ( ) ( ) ( )

( )) ( ( )( )

( )) ( ( )

(( ( )(

( )

( ) ( )

( ))

( ( )

)

( ))( )

x

f x g x xf x

f x g x x

f x x g xg x x

x xf x x g x

g x xx x

g x f x g x f x

f x

f x

f x f x

f x

g

xf x

xxx x g

g xx

1 1

Note that we add and subtract the blue expressions.Note that when 0, ( ) ( ).

Note that g (x) exists, and we can say that g(x) is continuous.

x g x x g x


( )uv u v uv


11 0

What is the derivative of the quotient of two differentiable functions?

( )Let ( ) , does ( ) exist?

( )

( )Note that ( ) !

( )

We use the definition of derivative:

( )( ) lim

x

f xh x h x

g x

f xh x

g x

h x x hh x

1

1 1 1 11 2

1

( )

Using the definition we can prove that:

( )If ( ) , then

( )

( ) ( ) ( ) ( )( )

( ( ))

The proof is similar to the proof of the product rule.

x

x

f xh x

g x

g x f x g x f xh x

g x


2

1 11 2 1

2 2

Extending the power rule to negative integers

If ( ) ( ) ?

1Let ( ) and use the qoutient rule:

(0)( ) ( )( )

( )

This shows that we

n

n

n n nn n n

n n

f x x f x

f xx

u u v uv

v v

x nx nxf x nx nx

x x

0

can extend the power rule to negative integers!

Remember that when 0 we use the constant function rule (x ) 0.

So far we have shown that we can use the power rule for all integers!

Later we'll show t

n

hat it can be used when n is a fraction,

and when n is a real number.


2

2

2

1( )

1

wh w

w

u u v uv

v v


0 0

0

0

0 0

Derivative of ( ) sin :

( ) ( ) sin( ) sin( )( ) lim lim

: sin( ) sin cosh cos sinh

sin cosh cos sinh sinlim

sin (cosh 1) cos sinhlim

sin (cosh 1) cos sinhlim lim

h h

h

h

h h

f x x

f x h f x x h xf x

h hNote x h x x

x x x

hx x

hx x

h

0 0

(cosh 1) inhsin lim cos lim

(sin )(0) (cos )(1)

cos

h h

hs

x xh h

x x

x

0

2 2

2 2

0 0

0

0

cosh 1 cosh 1: lim

cosh 1

Given: sin cos 1

cos h 1 sinlim lim

(cosh 1) (cosh 1)

sin sinlim 0

(cosh 1)

sinhlim 1 , So, (sin )

h

h h

h

h

Noteh

x x

h

h h

h h

h

and

dx

h dx

cos x


cos

sin .

Derivative of x is proved in a way similar to the way we proved

derivative of x Derivative of other trigonometric functions are found

by the Quotient Rule because we can write them as ratios of sin cos .x and x


2

2 22

2 2

(tan ) ?

sintan

cos

sin cos cos sin sin(tan )

cos cos

cos sin 1 = sec

cos cos

dx

dxx

xx

x x x x xx

x x

x xx

x x


2

2 2

2

(cot ) ?

coscot

sin1

Note: you could use cot to prove the result.tan

cos sin sin cos cos(cot )

sin sin

sin cos =

sin

dx

dxx

xx

xx

x x x x xx

x x

x x

x

2 22

2 2

(cos sin ) 1 csc

sin sin

x xx

x x


2

2

(csc ) ?

1csc

sin

1 (0) in (1)cos(csc )

sin sin

cos =

sin1 cos

csc cotsin sin

dx

dx

xx

s x xx

x x

x

xx

x xx x


2 2 2

(sec ) ?

1sec

cos

1(sec )

cos

(0)cos (1)( sin ) (0)cos (1)sin sin= =

cos cos cos1 sin

sec tancos cos

dx

dx

xx

xx

x x x x x

x x xx

x xx x


0

2 2

2 2

0 0

0

0 0

cosh 1 cosh 1: lim

cosh 1

Given: sin cos 1

cos h 1 sinlim lim

(cosh 1) (cosh 1)

sin sinlim 0

(cosh 1)

sinh coshlim 1 , So lim

h

h h

h

h h

Noteh

x x

h

h h

h h

h

and

h

10

h


2

( 4sin )( ) (4cos )(1)cos

x x xy x

x


Approximating using the derivative

ytany

x 1

1

tan

1

tan

tan

slope of the tangent line at

x x

x x

yx

x

y dy

x dx

dyy x

dx

tan

1

tan 1 1

1 1

1 1 1

1

1

t

by . This means thatwe can approximate )We ca

( ) by

( ) ( ). Note that is the tangent line

( ) ( )( - ).

We'll

n approxi

have:) ( ) ( )

mate (

(

y yy x f x

y y x x

f x

f x

y x y

y f x f x x x

x f x y f x y

1

an

tan1 1

tan1 11 1

1

1

) ( ) ( )

) ( ) ( ) ( )( - ).

(

(x x

dyx f x y f x xdx

f x

f x x f x y f x f x x x


3

3

2 2

tan1

3

1

(1.01) if ( ) .

0.01

( )

(1) 3 3(1) 3

(0.01) (3)(.01) .03

(1.01) (1) (1)(0.01) 1 0.03 1.03

Compare to (1.01) 1.030301 b

:

1

y calculato

r

x

Approximate f f x x

x

f x x

f x

Examp

d

le

yy

f

x

dx

f f

1

1

3 31 1

3 2 2 3 31 1 1 1

2 2 31 1

2 31

( ) ( )

3 3 ( ) ) ( )

3 3 ( ) ) ( )

3 ( ) ) (

:

)x x

x x

y x x x

y x x x x x x x

y x x x x x

dyy x x x x

dx

y dyk

x d

No e

x

t


Alternative proof of the product rule using linearization


1 1

0 :

Note: is the error whe

H

n we let

Note that is a variable and depends on .

ere we'll see that the err

Take the limit when 0 :

l

o

i

r goes to zero w

m

hen

x x x x

x

x

y dy y dyk k

x dx x dx

k x

x

1

1 1

1 1

1 1

0 0 0

0

0

0

0

lim lim

Note: limit of constant is constant: lim

lim

lim 0

lim 0

x xx x

xx x x x

xx x x x

xx x x x

x

y dyk

x dx

dy dy

dx dx

dy dyk

dx dx

dy dyk

dx dx

k


1

1

0

tan

tan

Given lim 0 and the equation below

we multiply both sides by :

So, we can approximate by

because 0 when 0

which makes (

x

x x

x x

k

x

y dyk

x dx

y dyx x k x

x dx

y y k x

y y

k x

k

) 0

But the reason approximation works even when is not

close to 0 is that ( )(the error) goes to zero faster than goes to 0.

See the figure above.

x

x

k x x


0

0

0

We know that

Given that lim 0

( ) ( ) ( ) Given that lim 0

( ) ( ) ( ) Given that lim 0

We can use this to approximate a function

xx a

x

x

dyy x k x k

dx

f a x f a f a x k x k

f a x f a x f a k x k

at

if we know ( ) and ( ). a x f a f a

Linear Approximation to ( ) at ( is a number close to )f x a x a


tan

3 2

Another result from this section is that

can be thought of as two quantities and where

(the y according to the tangent line)

So given ( ) , 3 can be written as 3

dydy dx

dxdy y

dx x

dyf x x x dy

dx

2 .

But, more importantly, we'll use this result

to discuss the Chain Rule next.

x dx

ytany

x


1

1 1

1

If is a function of and is a function of , then what is ?

t tx x

t t

t t

dxdydtdx

dy

dt

dyy x x t

dt

y x t

1 1 1

?

?

Is this always true?

( )( )

If ( ) and ( ), then

( ( ( )) ( ( )) ( )

t t x x t t

dy dy dx

dt dx dt

or

y f x x g t

f g t f g t g t


2

2

2 2

2

Example

( ) and ( ) 3 1

?

One way to find this is to find ( ) :

( ) (3 1) 9 6 1

18 6 18(2) 6 30

Can we do this without finding y(t)?

2 3(2) 1 5

2

t

t

y x x x t t

dy

dt

y t

y t t t t

dy dyt

dt dt

t x

dy

dx

5

2

2 5 2

2(5) 10

3 3

(10)(3) 30

The same answer! But does this formula always work?

x

t

t x t

dyx

dx

dx dx

dt dt

dy dy dx

dt dx dt

2( )y x x

( ) 3 1x t t

2( ) (3 1)y t t


1 1

?

If y is a function of x, and x is a function of t,

then y is a function of t.

But if ( ) exits and ( ) exists, then

does ( ) exist? If yes, how do we find it?

t t x x

y x x t

y t

dy dy dx

dt dx dt

1

is this true?t t

Differentiation of composite functions (the Chain Rule)


1

1

1

0

0

We start by the result we proved:

( lim 0)

We want , so we divide by :

( lim 0)

Take the limit of

x x

xx x

xx x

y dyk

x dx

dyy x k x kdx

dy tdt

y dy x xk kt dx t t

1

11 1

0 0 0 0 0

0

both sides when 0

lim lim lim lim lim

0 , ( lim 0)

t t t t tx x

xt tt t x x

t

y dy x xkt dx t t

dy dy dx dx kdt dx dt dt

dy dy dxdt dx dt

the desired result.


2 2

2 2 1 2

2 2 4 2

3 2

: ( 1) ?

If we use the power rule, we'll have:

2( 1) 2 2

But ( 1) 2 1

4 4 2 2

There is something wrong, and we are wrong

by a correction factor of 2 .

dyExample y x

dx

dyx x

dx

y x x x

dyx x x

dx

x

2

2

2

2 3

If we think of as a function of and as a

function of :

By the Chain Rule:1

(2 )(2 ) (2( 1))(2 )

(2 2))(2 ) 4 4

y u u

x

y u

u x

dy dy duu x x x

dx du dx

x x x x

The short cut is this ,

but we should not cancel the ' because

wh

:

en 0 , could equal zero.

y y x

t x t

x s

t x

Note


If , then ( ( ( )) ( ( )) ( )

If , then

f g x f g x f g x g x

dy dy duy u x

dx du dxExample

( ( ( )) ( ( )) ( )f g x f g x g x


( ( ( )) ( ( )) ( )f g z f g z g z

cos(4cos )( 4sin )

4sin cos(4cos )

z z

z z

0

0 0

0 0

0 0

Derivative of exponential function is:

( ) lim( )

( ) ( 1)lim( ) lim( )

( 1) (1 1)lim( ) lim( )

lim( ) lim(1)

x h xx

h

x h x x h

h h

hx x

h h

x x x

h h

e ee

h

e e e e e

h h

e he e

h hh

e e eh

1

0

1

1

Given that:

1lim(1 )

1

lim(1 )

(1 )

(1 ) )

(1 ) for small h

n

n

h

h

h

h hh

h

en

nh

h e

h e

h e

h e

(ln ) ?

ln

( ) ( )

( ) 1

1 1

y

y

y

y

x

y x

e x

e x

y e

ye x

Derivative of lnx:

( ) ?

ln ln

ln ln

1( ) ln

ln

ln

x

x

x

x

a

y a

y a

y x a

y ay

y y a

y a a

Derivative of : xa

log

(log ) ?

log

ln 1

1

ln1

ln

a

a

a

xy

y

y

y

x

y x

a a

a x

y a a

ya a

yx a

Derivative of : loga x

2

2

(log ) ?

1(ln ) (0)(ln )

ln

ln (ln )

1(ln )

ln

ln (ln )

ln 1

ln ln

1(log )

ln

a

a

x

a xx xa a

ax xa a

x

a x a

xx a

OR


0

0 0

1 1

1

1 11 0

00

0

00 0

0

If ( )

: ( ) ( ( ))

Thus: ( )

( ) ( )( ) ( ) lim

1 1lim = lim =

( ) ( )( ) ( ) ( )

y y

x x y y

y f x

Then f y f f x

f y x

f y f yf y

y y

x xf x f xf x f x f x

x x

0

10 0 0 0

0

0 0

Derivative of the inverse function

: x where is differentiable:

y ( ) and ( )

, since is continuous at :

, we have

Note At f

f x f y x

Also f x

as y y x x


3

3 3 3

1 21

3 32 233

Find the derivative of the inverse:

( ) 3

3 3 ( ) 3

This is the equation of the inverse without interchanging and .

1 1 1 1( ) ( 3) ( 3)

3 3 3 ( 3)3( 3)

Interchanging an

f x x

y x x y x y y

x y

x y y yyy

x

23

1 1

23

1 3

3

1d we get: ( ) ( )

3 ( 3)

1Letting ( ) ( ) ( ) ( ) .

3 ( 3)

But, using the formula for the derivative of the inverse:

1( ) ( ) Given: 3

( )

Note: ( 3) But, interchanging and

y y xx

y x f x f xx

f y y xf x

x y x

3

1 12 2 23 23

we get: y= ( 3)

1 1 1 1( ) ( ) ( ) ( )

3 3 3( 3) 3 ( 3)

y x

f y f xx y x x


1

3

( ) ( )-1

1

Find the derivative of the inverse at (4,1) on the graph of ( ) :

( ) 3

Using the formula for the derivative of inverse:

( 4 , 1 ) , (1,4) :

1( ) ( )

(

y coordOnf x x coordOnf x

f x

f x x

is on f so is on f

f yf x

2

12

1

) 3

1 1 1( ) (4)

(1) 3(1) 3

x

ff

1

:

The slope of the tangent line at

1(4,1) to f is .

3The slope of the tangent line at

(1,4) to f is 3.

Note

( )f x

1( )f x

1

1

1

1

1

2

(sin ( )) ?

( ) sin( )

we use the inverse

derivative formu

1( ) ( )

( )

la:

1

cos1

cos

1(sin (

(

)

) ( )

( )

)1

1

(

1

)

f yf x

f y

x

f x x

x

yf x

xx

x

Derivative of sine inverse function:2

1

1

Note: If this is easy to figure out.

But if sin ?

We set up the triangle below

to see the relation of :

Let sin an angle whose sine is x.

Note: this y is the inverse, ( )

y x x y

y x what is x

x and y

y x

f x

, not ( ).f x

y

x

21 x

cosadjacent

yhypotenuse


1

2

sin ( )

1( )

1 ( )

y something

y somethingsomething

1 3sin ( 3 )y x

Derivative of cosine inverse function:

1

1

2

1

1

1

1( ) ( )

( )

( ) ( )

(

(cos ( )) ?

( ) cos( )

1

sin1

sin

1(cos ( ))

11 1

) ( )

f yf x

f y

f

x

f x x

x

y

xx

x

x

y

x

21 xsin

oppositey

hypotenuse

1

In the triangle:

cos an angle whose cos is xy x


1

2

cos ( )

1( )

1 ( )

y something


1 5cos ( 3 )y x

y

Derivative of tangent inverse function:

x

21 x

1

1

1

1

1

2

2

2

(tan ( )) ?

( ) tan

1

(sec )

1

(sec )

1( ) ( )

( )

( ) ( )

1

1

( ) ( )

( ) ( )

( , )

x

f

f yf x

f y

x x

x

y

xx

f x

f x

2

2 2

1

2

tan an angle whose tan is x

1cos

1

sec 1

sec 1

yx

y x

y x

y x

2

2

1cos

1

sec 1

yx

y x


1

2

tan ( )

1( )

( ) 1

y something


1 3tan ( 4 )y x

Derivative of cotangent inverse function:

1

1

2

2

1

1

2

1 1

(cot ( )) ?

( ) cot

1

( csc )

1

( csc

( ) (

)

1

1( , )

)( )

( ) ( )

( ) ( )

( ) ( )

x

f x x

x

y

xx

f yf x

f y

f x

f x

2

2 2

1

2

cot an angle whose cot is x

1sin

1

csc 1 csc 1

yx

y x

y x

y x

y

1

x

2 1x

2

2

1sin

1

csc 1

yx

y x


1

2

cot ( )

1( )

( ) 1

y something


1 3cot ( 2 )y x

1

1

1

1

2

1

2

1

(sec ( )) ?

( ) sec

1

(sec tan )

1

(sec tan

1( ) ( )

( )

)

1

1

1(

( ) ( )

( ) ( )

( ) ( )

sec ( ))1

1

f yf x

f y

f x

f

x

f x x

x x

y y

x x

xx

x

x

x

Derivative of secant inverse function:1

2

sec an angle whose sec

1cos sec

tan 1

is x

y y xx

y x

y x

y

x

1

2 1x

2

1cos sec

tan 1

y y xx

y x


1

2

sec ( )

1( )

( ) 1

y something

y somethingsomething something

1 3sec ( 3 )y x

1

1

1

2

1

2

1

1

1( ) ( )

( )

( ) ( )

(

(csc ( )) ?

( ) csc

1

( csc cot )

1

( csc cot )

1

1

1(csc ( ))

1

1

) ( )

( ) ( )

x

f x x

x x

y

f yf x

y

x x

xx x

x

f y

f x

f x

Derivative of cosecant inverse function:1

2

csc an angle whose csc is x

1sin csc

cot 1

x

y y

y

y

xx

x

y

x

1

2 1x

2

1sin csc

cot 1

y y xx

y x


1

2

csc ( )

1( )

( ) 1

y something

y somethingsomething something

1

55

4

5

( ) ( )

1( ) ( )

5

y something something

y something something

1 25 csc ( )y x


Derivative of ( )

ln ln

1 1(1) ln ( )

ln 1

(ln 1)

(ln 1)

x

x

x

f x x

y x

y x x

y x xy x

yx

y

y y x

y x x


8 6 4 6

Implicit Differentiation

Given ( ) we can always find ?. The assumption we

make is that we can explicitly solve for .

What if we are given the following, and ask to find ?

x 3

dyy f x

dxy

dy

dx

x y y


8 6 4 6

2

2

x 3

Note: There is a difference between an equation

and an identity: 0 is an equation that is

true for ( , ) such that ( , );for example (3,9) works.

But an identity is true for all val

x y y

x y

x y x x

2 2 2

2

ues of the

variables and it is represented by instead of =.

( ) 2 is true for all ( , ).

Or, 1 ( 1)( 1) is true for all values of x.

When we take the derivative of both sides of an id

x y x xy y x y

x x x

2

enty

we get an equality.

( 1) [( 1)( 1)] 2 2x x x x x


Assume ( ) such that when we substitute ( ) into:

1 for we get an identity.

1Let , then 1

for all values of except 0.

Since we are assuming that y is a function of x we can use

the pro

y f x f x

xy y

y xyx

x x

2

2

duct rule to take the derivative of both sides of

( ) (1)

( ( )) ( ) 0

11

0

1 1But if , then we can compute directly:

d dxy

dx dxd d

x y y xdx dx

dy dy dy y xy x x ydx dx dx x x x

dyy

x dx x


2 2

2 2

: Assume that ( ) such that 25

2( ) (25) 2 2 0

2

If we now differentiate directly we get the same answer:

Example y y x x y

dy dy x xx y x y

dx dx y y

dy x

dx y

2 2

2

2 21 2

121 2

12 12

122 2

1 12 2 22 2

Afetr solving for in 25 :

25 which means we have two functions:

y 25 , y 25

1 2(25 ) ( 2 )

22(25 )

1 2(25 ) ( 2 )

22(25 ) (25 )

y x y

y x

x x

dy x xx x

dx yx

dy x x xx x

dx yx x


8 6 4 6

8 6 4 6

8 6

Find the equation of the tangent line to

3 at the point (1,1).

We assume that ( ) is a function of x such that

if substituted for y it would make the equation an identity:

3

(

x x y y

y f x

x x y y

x x

4 6

7 5 4 6 3 5

7 5 4

6 3 5

(1,1)

) (3)

8 6 4 6 0

(8 6 )

4 6

14 7

10 5

71 ( 1) 7 5 12

5

y y

dy dyx x y x y y

dx dx

dy x x y

dx x y y

dy

dx

y x x y


6(0, 3)

6(0, 3)

8( 3,0)8( 3,0)

(1,1)( 1,1)

( 1, 1) (1, 1)

8 6 4 6

Note that given:

3

If we were told to find the tangent line to the graph at 1, there

would be difficult to figure out what the value for 1 is!

We would have to solve a 6th degree polynomi

x x y y

x

y x

8 6 4 6

al!

1 1 3y y


Note: The previous problem was "cooked" so that we could

graph it, and we were given points with x and y coordinates

known. In general, it is very difficult to graph these

equations. For example, we can tell from the graph that

there are two tangent lines at 1, and two tangent lines

at 1.

x

y

(1,1)( 1,1)

( 1, 1) (1, 1)

This is why we need to know where

the equation has vertical tangent lines

because around these points the graph

becomes a multi-valued curve. We can

also assume that both and are

functions of another

x y

variable, namely, t.


Use implicit differentiation to find in terms of and .dy

x ydx

3 23 2x y 2

2

9 2 0

9

2

dyx y

dx

dy x

dx y


2 2

3, 4

2 2

3, 4

Relate Rates

A particle moves along the circular path 25.

( and in feet). At (3,4), 8 sec

Find ?

( ) (25)

2 2 0

2

2

3(8) 6 s4

x y

x y

x y

dx ftx ydt

dy

dt

d dx y

dt dtdx dy

x ydt dt

dxxdy x dxdt

dt y y dt

dy ftdt

ec

dx

dt

dy

dt


A hypothetical square grows so that the length of its sides

are increasing at a rate of 9 m/min. How fast is the area

of the square increasing when the sides are 14 m each?

2

2

14

2 2(14)(9) 252mins

A s

dA ds dA ms

dt dt dt


A 17 ft ladder is leaning against a wall and sliding toward

the floor. The top of the ladder is sliding down the wall

at a rate of 6 ft/sec. How fast is the base of the ladder

sliding away from the wall when the base of the ladder is

15 ft from the wall?

dy

dtdx

dt

2 2 2 2 2 2

2 2 2

: 17 15 17 15 64, 8

( ) 17 2 2 0

Note x y x y y

d d dx dyx y x y

dt dt dt dt


dr

dtdV

dt

3 2

2 210

44

3

15 30.012

4 4 10 80 min minr

dV drV r r

dt dtdV

dr dr in indtdt r dt


250 Given: 30sec

50 2

10 50(30) 2(10)(30) 900sec

40 50(30) 2(40)(30) 900sec

dx fty x x

dtdy dx dx

xdt dt dt

dy ftx

dtdy ft

xdt


Higher-order Derivatives

1 1 1( 1) ( )2 2 2

1 3( 1)2 2

52

52

5

Find the third derivative of ( ) .

1 1( ) ( )2 2

1 1 1( ) ( )( )2 2 4

1 3 3 1( ) ( )( ) ( ) ( )4 2 8

3 1( ) ( )8

f x x

f x x f x x x

f x x x

f x x f xx

f xx

Geogebra’s answer


(1)

2(2)

2

3(3)

3

4(4)

4

( )

Notation for higher derivatives:

( ), , y , ( )

( ), , y , ( )

( ), , y , ( )

.........., ,....., ( )

.........., ,....., ( )n

nn

dyf x f x

dx

d yf x f x

dx

d yf x f x

dx

d yf x

dx

d yf x

dx

(0): ( ) ( )Note f x f x


Tangent lines and Normal linesThe normal line to a curve C at a point A is the line through A that is perpendicular to the tangent line at A.

The concept of the normal line to a curve has applications in the study of optics where one needs to consider the angle between a light ray and the normal line to a lens.


2

Find equations for the tangent and normal lines to the graph

of ( ) at (2,4).

( )

f x

f x x

( ) 2 (2) 4

: (2) (2)( 2) 4 4( 2)

1 1: (2) ( 2) 4 ( 2)

(2) 4

f x x f

Tangent y f f x y x

Normal y f x y xf

calculus i chapter three1. 2 calculus timeline: descartes 1596-1650 cavalieri 1598-1647 fermat...

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