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

- anonymous

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- amistre64

how bad can it possibly be?

- anonymous

just not sure what to do?

- amistre64

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

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## More answers

- amistre64

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

- anonymous

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

- amistre64

its a few chain rules

- anonymous

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

- amistre64

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

- amistre64

53-56 is about the graph

- amistre64

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

- anonymous

yeah just too clarify

##### 1 Attachment

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

h(1) = f(g(1))
whats g(1) = ?

- amistre64

54 you say .... that one was 53

- anonymous

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

- amistre64

u = g(f)
u' = g'(f) * f'
its just a chain rule

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

56 is same idea as 54
define your chain rule and plug in slopes and values etc ....

- anonymous

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

- amistre64

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

- anonymous

deal

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

yep

- amistre64

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

- amistre64

4 up and 2 over is 4/2 = 2

- anonymous

u r right did it the wrong way :)

- amistre64

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

- anonymous

correct

- amistre64

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

- anonymous

so u=g(f)
u'= g'(f) * f'
u'= -1 * 4
= -4

- amistre64

when is the slope of f' = 4?

- anonymous

2

- amistre64

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

- anonymous

Yes it is but I am so confused...

- amistre64

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

- amistre64

x<2; f'=2
x>2; f'= -2
BUT, at x=2, we have no way to define the slope of a pointy place.

- anonymous

there is no slope there so no derivative?

- amistre64

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

- amistre64

so u'2 cannot be defined either

- anonymous

ok let me try 3

- anonymous

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

- amistre64

exactly :)

- anonymous

GREAT!!! Thank you!!!

- anonymous

so i think I can do 56 what about 58

- amistre64

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

- amistre64

D(f(g)) = f'(g) * g'
define your slopes for given xs and use that to determine the answer

- anonymous

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

- amistre64

your welcome :) and good luck

- anonymous

thx

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