Question

Derivative help please!!!
http://www.wiley.com/college/sc/hugheshallett/chap3.pdf
Page 131 number 52 and 54

Answer 1

how bad can it possibly be?

Answer 2

just not sure what to do?

Answer 3

\[ g(a) = ae^{-2a}\]
this one?

Answer 4

\[g(a) = e^{ae^{-2a}}\]

Answer 5

e^ae^-2a
e^ae dx ae^-2a
e^ae dx ae^-2a dx -2a

Answer 6

its a few chain rules

Answer 7

no on the pdf I posted page 131 numbers 52 and 53 about the graph

Answer 8

yeah, thats the one i posted, after trying to find the needle in a haystack ....

Answer 9

53-56 is about the graph

Answer 10

i would create a few linear equations to help out with 53-56

Answer 11

yeah just too clarify

Answer 12

lol, that ones alot easier to find stuf on :)

Answer 13

should have done that to start with lol.... I need most help on 54, 56,and 58

Answer 14

and since the graph is lines to begin with , the derivatives just define the slopes of the lines

Answer 15

h(1) = f(g(1))

whats g(1) = ?

Answer 16

54 you say .... that one was 53

Answer 17

not sure.. I think I have confused myself
yeah just 54, 56, and 58

Answer 18

u = g(f)

u' = g'(f) * f'

its just a chain rule

Answer 19

so; u'(1) = g'(f(1)) * f'(1)

Answer 20

then the same concept for u'2 and u'3

Answer 21

ok cool thanks... I guess I was on track just late i guess lol
what about the 56, 58

Answer 22

56 is same idea as 54

define your chain rule and plug in slopes and values etc ....

Answer 23

will you finish going through u'(1) with the numbers so I can see it all the way through

Answer 24

ill guide you, but youve got to find the numbers :)

Answer 25

deal

Answer 26

for example; f'(1) means the slope of f at x=1

so it might be a good idea to define the slopes of the 3 parts of the graph for good measure

Answer 27

whats the slope of f on the left?, of f on the right? and of g?

Answer 28

slope of f on left is 1/2 and right f is -1/2?
slope of g is -1?

Answer 29

yep

Answer 30

or is it 2 and -2 ..thats seems better

Answer 31

4 up and 2 over is 4/2 = 2

Answer 32

u r right did it the wrong way :)

Answer 33

so: g' = -1 no matter what x is
f' = 2 when x<2
f' = -2 when x > 2

agreed?

Answer 34

correct

Answer 35

u = g(f)

u' = g'(f) * f' ; when x=1,2,3 we just plug it in

u'1 = g'(f1) * f'1
= -1 * 2
= -2

do x=2 and x=3 for me based on this

Answer 36

so u=g(f)

u'= g'(f) * f'

u'= -1 * 4

= -4

Answer 37

when is the slope of f' = 4?

Answer 38

2

Answer 39

at x=2 ? is that your guess :) i already know why if so but its incorrect

Answer 40

Yes it is but I am so confused...

Answer 41

did we define the slope for f' at x=2?

Answer 42

x<2; f'=2
x>2; f'= -2

BUT, at x=2, we have no way to define the slope of a pointy place.

Answer 43

there is no slope there so no derivative?

Answer 44

correct :) no definable way to determine a derivate at a cusp, corner, pointy place

Answer 45

so u'2 cannot be defined either

Answer 46

ok let me try 3

Answer 47

g'(f(3))*f'(3)
g'(2)(-2)
(-1)(-2)
2????

Answer 48

exactly :)

Answer 49

GREAT!!! Thank you!!!

Answer 50

so i think I can do 56 what about 58

Answer 51

58 is the same thing; define your slopes from the given graphs to aid you in finding a solution to what they are looking for

Answer 52

D(f(g)) = f'(g) * g'

define your slopes for given xs and use that to determine the answer

Answer 53

ok I think I can do that from the example above... it clicked in that last example... thx again for you help

Answer 54

your welcome :) and good luck

Answer 55

thx