-cItrAIN

RULE-

il

-Basic

idea:

Multiplying and/or

dividing certain

derivatives

in order

to obtain

a desired derivative.

-There

are several forms

of

the chain rule.

l)

Multiplying derivatives:

This

form is

widely used to expand

our formula

base.

It is

sometimes called

the

composite form,

since

it is

used

for

derivatives

of functions

"within"

functions.

See the calculus

web

site link

on my home

page

for

a very nice

discussion.

-Note:

This is its form:

If h(x):g[(x)],

then

h'

(x;:g

'[f(x)]

f

'

(x)

2) Dividing

derivatives:

This fonn

is widely

used for

derivatives

of

paramehically

defined

functions.

These

are curyes

where

a

point

(x,y)

is determined

by

a 3rd

quantity (parameter).

So, x

and

y

are

given

in

terms

of this

parameter.

Popular

parameters

are

time t and angles

0

(theta).

Values

for the

parameter

will determined

the

location

of the

point (x,y)

on our curve. In calculus

II and

III,

you

will

study

vector

functions

which

will have

parametric

components.

This

gives

a

point

much

freedom

of motion

since the

curve is

not required

to be

a function

of

x.

(the

coordinates

x

&

y

are firnctions

of the

parameter).

A

popular

application

in

calculus

III would

be the analysis

of

space

curves.

This

form

also allows

us

to solve interesting

rate

of change

problems.

(finding

rate

of change

of one

quantity

with

respect

to another

quantity).

Here is

an example

of both the

parametric

form

and the

rate

of

change application

using farm#2

of the chain

rule.

Example:

Given

parametric

equations

x

:

e

o'

and

y

:

ln(t3

+2t+7),

find

dy/dx.

Note

that dx/dt:

eo'(4)and

dy/dt

:

lll(f

+2t+7)7(3t2

+z)are

easily found.

To

get

dy/dx,

we divide

(dyldt)

by

(d/dQ.

We

ffeat

the derivatives

as

if

they were

fractions

with

sepmate numerators

& denominators

ltney

are not,

not

yet..we

will define

dx and

dy separately,

coming

soon...they will

be called

differentials).

So, for

this

problem,

dy/dx:

{[1(

f

+2t+7)](N2

+2)]l

eo,

(4):

(312

+Dfi4

ea,

7t,

+Zt+711

Example:

Find

the

instantaneous

rate

of change

of

the volume

of an expanding

cube

with respect

to

(wrt)

its

surface

area

at the instant

when

the edge is

3

inches.

The

formulas

here

are v:

x3 and S

:6x2,

where

x represents

the edge

of the

cube.

We

want dV/dS

at x:3".

We

can easily

get

dV/dx:3x2

and

ds/dx

:

l2x,so,

dV/dS

=3*

ll2x:

e/$x.

At

x

:

3, dV/dS

:314

rnches

cube

per

square inch

*Note:

Using

this

technique,

we can find

the instantaneous

rate

of change

of any quantlty

wrt

another

quantity.

Just form

two

equations

with

the letters

of

your

choice

by calling

each

a single

variable.

Example^:

Find

the rate

of

change of

x2

+

x

-7

wrt

x /(x-l)

Let

s

:

*'

+

x -7

and.g

:

x

/(x-1)

then

proceed

as in the last

example.

Your

letters

could

be much

more

creative.

(compute

ds/dg)