Bear H.S. Understanding Calculus

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Chapter 3 • Differentiation

3.3

f(x)

= x

;

a = 3

Hint: x

= (x - a)(x

+ax +a

) .

3.4

f(x)

= 2x

;

a = 1

3.5

f(x)

3JX;

a = 9

3.6

f(x)

-JX

+x;

a = 1

3.7

f(x)=-;a=3

3.8f(x)=2";a=1

3.9 f

(x)

=x j ; a =8

1 1 2 1 1 2

Hint: (x

aj)(x

ajx

x-a.

Find the equation of the line tangent to the curve at the given point.

3.10 y = 2x

;

(1,2)

3.11 y = 5 - x

; (2, 1)

3.12 y =

2JX;

(4, 4)

3.13 y =

JX;

(1, 1)

3.14 A ball is thrown straight up from the ground, and its height s in feet after t seconds is

lOOt

- 16t

•

What is the speed of the ball at t = 2 seconds? What is the initial speed

s'(O)?

How high does the ball go?

(Hint:

The ball reaches maximum height when its speed

is zero.)

3.15 A ball thrown down from the top of a 104-ft building, with an initial speed of 20 ft/sec,

travels a distance s

= 20t + 16t

in t seconds. When does it hit the ground, and how fast

is it going then?

3.16 A spherical balloon is blown up in such a way that its radius at

t seconds is t em; that is,

r = t. What is the rate of change of the volume at any time t? The volume is V =

rr,3 ,

and the surface area is S

= 4rr,2. How do you interpret the fact that

= S?

3.17 A block of ice initially weighs 100 lbs and melts so that its weight

W after t minutes is

100 -

20.

At what rate (lbs/min) is the ice melting at t = 100 min?

Differentiation

Formulas

In this section we develop some general rules for differentiating functions, so we

don't

have

to go through the limit argument each time.

What

we have so far are techniques that will give

us the derivative of a specific function at a specific

point-for

example, the derivative of x

= 1. Now we search for a formula for the derivative

at any x, and

course we want

such formulas for all the common functions.

Our

definition for

(xo) is the following:

'( )

1·

f(x)

f(xo)

x~xo

X -

(4.1)

(4.2)

To get a formula for

f'(x),

for any x, we replace

in (4.1) by x, and write the nearby point

x +

~x.

Here

just

stands for a small change in x. With this notation,

f(x)

f(xo)

becomes f (x +

~x)

- f (x), and x -

becomes

~x.

This gives us the alternative form

'( ) -

1·

f(x

~x)

f(x)

x -

~x~O

If Y =

f(x),

we let

f(x

~x)

!(x),

is the change in y that corresponds to

the change

in x. The difference quotient at any point x is

f(x

~x)

f(x)

We write

for the limit of

0, so we have the following alternative notations:

= lim

!:J.y

= lim

f(x

!:J.x)

f(x)

!,(x).

~x~O

We also use

as a differentiation operator so that

f(x)

f'(x).

In the examples of the last section, we saw that to differentiate the sum

two terms you

just

differentiate the terms separately and add the results. Thus, if y =z +

where z and W

are functions of x, then

~w,

(4.3)

Understanding

Calculus

and consequently

-=-+-,

dy = lim

(~z

~w)

~x-+o

1·

=lm-+lm-

~x-+o

The derivative of a sum (or difference) is the sum (or difference) of the derivatives. In the

second equation above we used the fact that the limit of a sum is the sum of the limits. This is

an obvious property of addition: If

a is close to A and b is close to B, then a + b is close to

A + B. Similar statements hold for a - b, ab, and a/ b, so, for example, the limit of a product

is the product of the limits, and so on.

Now consider the derivative of a product of two functions. Let

y =z. w, and again let

~z,

be the changes in z and w that result when x is changed by an amount

~x.

The new

value of y is (z +

~z)(w

~w),

and therefore

= (z +

~z)(w

~w)

- zw

= z:

ui-

~z·

~w.

It follows that

~W)

lim

z-+w-+~z·-

~x-+o

=z-+w-+O·

=z-+w-.

Of course

lim~x-+o

az =0, for otherwise

~z/

would not approach a limit as ax

function z with the property that

0 as

0 is continuous at the point in question.

Differentiable functions are continuous, and calculus deals with differentiable functions.

In functional notation formula (4.3) can be written as follows: if

f(x)

g(x)

h(x),

then

f'(x)

g(x)h'(x)

h(x)g'(x),

d d d

f(x)

g(x)

h(x)

dxg(x).

The derivative (rate of change) of a constant function is obviously zero:

It is

also clear from the definition that

d dy

(cy)

The derivative of the function x is clearly one

(~~

= 1), so

(cx)

= c.

-1

Chapter

4 •

Differentiation

Formulas

Now

we use the product ruleto

find

formula

for

•

d 2 d dx dx

- x

= - (x . x) =x - + x - =2x,

dx dx dx dx

d d d dx

- x

= - (x . x

)

= X- x

+ x

= x . 2x + x

=3x

dx dx dx dx

d d d dx

-(x.

)

x-x

x·

=4x

•

dx dx dx dx

The patternis

clear:

_xn=nx

n=O,I,2,3,

....

follows

that

EXAMPLE 4.1

Find

if y = 5x

- 3x

+ 2x

- 5.

Solution

Differentiating the termsseparately and

adding,

we get

- =50x

15x

+ 4x.

find

formula

for

;xx-

, we startwith

x-I

and

find

d~;l

fromthe

definition:

I I

d 1

x+~x

- x

dx x

~x~O

x - (x +

Ax)

~x~o

x(x

~x)

= lim

~x~O

x(x

+ dX)

-1

- x

That is,

(4.4)

-1

-2

-x

. (4.5)

Now proceed as before, using the product rule and (4.5):

d d

dx-

dx:'

_x-

_(x-

.x-

)

=x-

_ -

+x-

dx dx dx dx

= X-I

(_x-

)

x-I

(_x-

)

-2x-

d d

dx-

-x-

_(X-I.

)

x-I

dx dx dx dx

I(-2x-

)

+x-

2(_x-

)

-3x-

•

Similarly,

we show that

-4

-x

dx '

-5

-x

= - x ,

Understanding

Calculus

and so on, and the general formula is

-n

-n-l

-x

-nx

1,2,3,

....

(4.6)

Notice that the rule in (4.6) is the same as that for positive exponents: to differentiate x", for n

positive or negative, multiply by the exponent, and subtract one from the exponent to get the

new exponent.

EXAMPLE

4.2

Find

if y = 5 - L + 2x

•

dx . ;J x

Solution

Write the expressionwith negative exponentswhereappropriate, and use (4.4) and (4.6).

y =

5x-

7x-

+ 2x

dy =

-15x-4

14x-

+ 8x

•

To find a formula for the derivative

the quotient

two functions, dd

(.£),

we first find

x W

(1),

and then we can use the product rule to find dd (.£) = dd (z .

1).

If y =

then

x w w

Ay 1

1 ]

w +

w - w

=_1

[w-(w+~w)]

w(w +

~x)

-1

w(w

~x)

Hence, taking the limit as

dy _

(~)

1 dw

w - w

Now we use the product rule to find

(

(4.7)

(4.8)

(~)

= !£

z!£

(~)

w w

=z(-I)dW+~dZ

_Zdw

dx dx

=-----

_ Zdw

dx dx

=----

This formula is best remembered as "the bottom times the derivative

the top, minus the top

times the derivative

the bottom, over the bottom squared," for this chant

doesn't

depend on

which variables are used for the numerator and denominator.

EXAMPLE

4.3

Find

if y = x

3+3x.

dx x

Chapter 4 • DifferentiationFormulas

Solution

dy _ (X

+ 2)-j;(x

+ 3x) - (X

3x)-j;(x

+ 2)

dx - (x

+2)2

+ 2)(3x

+ 3) - (X

+ 3x)2x

(x + 2)2

+9x

+ 6 - 2x

(x + 2)2

=(X

+ 3x

6)/(x

+ 2)2.

(4.9)

EXAMPLE

4.4

Find *if y = (x

+ 1)(4x - I).

Solution

We can either use the product rule, or first multiply the two terms and differentiate the resulting sum.

With the product rule we get

- = (x

+ I)

·4+

(4x - 1)2x

=4x

+4 + 8x

=12x

- 2x + 4.

If we multiply out first, we have

y = (x

+ 1)(4x - I) = 4x

+4x - I,

- = 12x

2-2x+4.

To add some variety to our repertoire,we now compute

fx.JX:

Jx+~x-.JX

-yX

= im

~x~O

= lim

(../x

- .JX)(Jx+

.JX)

~x~O

~x(../x

+.JX)

x +

~x-x

= 1m

~x~O

(Jx +

+.JX)

1 1 I

-X-

2.JX

- 2 .

Notice that the formula above,

lxx! = !x-!, follows the same rule as for integer exponents:

= nx

•

EXAMPLE

4.5

· d

";[;

3x-2'

Solution

!£ y'X _ (3x - 2)

- y'X . 3

dx 3x - 2 - (3x - 2)2

3x - 2 - 6x

2y'X(3x - 2)2

2y'X(3x - 2)2.

Understanding

Calculus

PROBLEMS

Find

4.1 y = Sx

+ X

2 1

4.2 y =

-3x

- 2

4.3 y =

3x-

X-I

+ 2x

2 3 1 6

4.4 y = - - -

+Sx

+-x

3 1

4.5

y=x

10 1

4.6 y

4.7 y = (x

+ 1)(2x + 3)

4.8 y =(x

+1)(x

+2x

4.9

y=----

4.10 y = x

+ 1

4.11

y = -2-+-x-

4.12 Y =

x-+-l

4.13 y =

-2-

X +1

2x +1

4.14 y = -2x---l

4.15 y =

s,JX

+ -

4.16 Y = JX(3x - 2)

4.17 Y =

-JX-x

4.18 Show that formula (4.9) for

1;x!

and your answer to Problem 4.17 for

1;x-~

both agree

with the rule for integer exponents:

= nx

•

4.19 Show that .!L(u .

z·

w) =

zw~

+ uw4:£ + uz

•

dx dx dx dx

In Problems 4.20 and 4.21 use the formula of Problem 4.19 to find

4.20 y = (x + 1)(x

+2)(x - 2)

4.21 y =

3(x

3)(x

4.22 Check that

1;x-

-4x-

by differentiating the product:

(x-

) .

4.23 Check that

1;x-

-5x-

by differentiating the

product

(x-

•

) .

4.24 Find

1;x!.

Hint: Use (a - b)(a

+ab + b

)

= a

•

Multiply the top and bottom of

I 1

[(x+~1:

-x

by a

+ab +b

with a = (x +

~x)!

, b =

4.25 Use the product rule to show that if N is a positive integer and

!;x

= NX

then

!;(x

•

x) =

l)x

•

This is the essential step in the inductive proof that

/;x

=nx

for all positive integers.

The Chain Rule

Most of the functions we use are composites of simpler functions. For example, (2x

+ 1)3

is calculated by evaluating 2x

+ 1 and then cubing that number. If

f(x)

= 2x

+ 1 and

g(x) = x

then

g(f(x»

f(x)3

= (2x

+ 1)3.

Similarly, we can write

,J3x

+ 1 as the composition

..jX

and 3x +1: if

f(x)

=3x +1 and

g(x)

..jX,

then

g(f(x»

= J

f(x)

,J3x

+ 1.

Here are some other examples:

1 . 1

(i)

-2-

g(f(x»

with g(x) =

f(x)

= x

+4;

(ii) (3x - 1)4 =

g(f(x»

with

g(x)

= x

f(x)

= 3x - 1;

1 1

(iii)

.JX2+X

h(g(f(x)))

with

h(x)

g(x)

,JX,

f(x)

= x

+x.

Using dependent variable notation

(y,

etc.) instead of functional notation

(f(x),

g(x),

etc.), the above examples would look like this:

') .

f 1 d 2 4 h 1

1 Z = - an y = x +

en z =

-2--;

y x

(ii) if z = y4 and y = 3x - 1, then z = (3x - 1)4;

(iii) if z = 1 and W = If and y = x

+ x, then z =

~r~

Situations like these are quite common in physical applications. For example, if the force F

on an object depends on its position x, and the position x depends on the time t, then F is a

(composite) function of

t. Because most functions that occur in practice are compositions

simpler functions, we proceed to find the rule for differentiating such functions.

Consider the general situation in which z is a function of

y and y is a function of x, so

that z also becomes a function of

x. Let

t!,.y

be the change in y which results from changing x

(5.1)

Understanding

Calculus

by an amount

Sx,

Let

be the changein zwhichresultsfromchangingy by

y, so

also

corresponds to the change

in x. Noticethat

!::"y

0 as

!::"x

0, since y is necessarily

continuous if

exists. Hence,

dz = lim

!::,.z

~x-+o

= lim

!::,.z.

!::"y

~x-+o!::"y

. dz dy

= lim

-.-.

~x-+o

dy dx

This is the importantchain rule for differentiating compositefunctions.

In the argumentleadingto

(5.1), we multipliedand dividedby

!::"y,

which is clearlynot

legitimateif

!::"y

=0 for arbitrarily small values of

Sx,

However,

in that case

!::,.z

is also zero

for arbitrarilysmallvaluesof

!::"x,

sincezdoesn't changeif y doesn't change, and both

and

wouldthen be zero, and (5.1) would still hold in the form 0 =

Weillustratethechainrulewiththefunctionz = (2x

+1)3, or z = y3 with y =2x

+ 1.

From (5.1),

dz dz dy

= dy . dx

y2(4x)

= 3(2x

+ 1)2(4x).

In practice,we think of this computationas

( )3 =3( )2

(

whichholdsnomatterwhatfunctionis insidethe parentheses. Forexample,if z =

(.JX

) 3,

then

dz =3(

)2!!-(

)

dx dx

(JX

)2 •

!!-

(JX

)

(JX

) 2 .

(2~

) .

Toillustratethe chain rule with a differentoutside function, consider y = 1/(3 - x

)4 .

Here

y = (

)-4,

dy 5 d

- =

-4(

)

dx dx

2 5 d 2

-4(3

- x )- - (3 - x )

-4(3

- x

2)-5(-2x).

=----

Chapter 5 • The Chain Rule

Here are some more examples.

~(3x

- 6x

)5 =5(3x - 6x

)4 • (3 - 18x

2);

-(5x

+ 1)-2 =

-2(5x

+ 1)-3 . (15x

) ;

dId

---

-(1

)- 1 =

-(1

2)-2(2x).

Recall that

(,Ji)

21,

so the chain rule gives the general formula

dId

J()

( ) dx ( ).

For example,

~,/3

- 5x = 1

(-5);

2J3

- 5x

~Jx2

+ 5x

= 1 (2x + 20x

) ;

2Jx

+ 5x

1 (1)

dx V1+

2)1

- x

Now consider z =

Jl~X2.

This is a composition

three functions:

1 2

z=-,w=,J"Y,y=l+x.

The derivative is

d: dz dw dy

= dw . dy . dx

1 1

=----·2x

1 1

= - .

·2x

(1 + x

)

2.Jf+X2

=,-x/(1

+x2)~.

We could also find

~.~

by the quotient rule in this example.

d 1

.Jf+X2.!L(I)

- 1 .

.!LJl

+ x

dx dx

---

-----------

.Jf+X2

(1 + x

--1-(2x)

2Jl+x

(1 + x

-x

EXAMPLE

5.1

Grain pours out of a spout at 5 cu ft/min onto a concrete floor and forms a conical pile whose radius and

height are equal. At what rate is the height of the pile increasing when

h = 4

£1?