anonymous
  • anonymous
Derivative help please!!! http://www.wiley.com/college/sc/hugheshallett/chap3.pdf Page 131 number 52 and 54
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
how bad can it possibly be?
anonymous
  • anonymous
just not sure what to do?
amistre64
  • amistre64
\[ g(a ) = ae^{-2a}\] this one?

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amistre64
  • amistre64
\[g(a) = e^{ae^{-2a}}\]
anonymous
  • anonymous
e^ae^-2a e^ae dx ae^-2a e^ae dx ae^-2a dx -2a
amistre64
  • amistre64
its a few chain rules
anonymous
  • anonymous
no on the pdf I posted page 131 numbers 52 and 53 about the graph
amistre64
  • amistre64
yeah, thats the one i posted, after trying to find the needle in a haystack ....
amistre64
  • amistre64
53-56 is about the graph
amistre64
  • amistre64
i would create a few linear equations to help out with 53-56
anonymous
  • anonymous
yeah just too clarify
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amistre64
  • amistre64
lol, that ones alot easier to find stuf on :)
anonymous
  • anonymous
should have done that to start with lol.... I need most help on 54, 56,and 58
amistre64
  • amistre64
and since the graph is lines to begin with , the derivatives just define the slopes of the lines
amistre64
  • amistre64
h(1) = f(g(1)) whats g(1) = ?
amistre64
  • amistre64
54 you say .... that one was 53
anonymous
  • anonymous
not sure.. I think I have confused myself yeah just 54, 56, and 58
amistre64
  • amistre64
u = g(f) u' = g'(f) * f' its just a chain rule
amistre64
  • amistre64
so; u'(1) = g'(f(1)) * f'(1)
amistre64
  • amistre64
then the same concept for u'2 and u'3
anonymous
  • anonymous
ok cool thanks... I guess I was on track just late i guess lol what about the 56, 58
amistre64
  • amistre64
56 is same idea as 54 define your chain rule and plug in slopes and values etc ....
anonymous
  • anonymous
will you finish going through u'(1) with the numbers so I can see it all the way through
amistre64
  • amistre64
ill guide you, but youve got to find the numbers :)
anonymous
  • anonymous
deal
amistre64
  • 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
  • amistre64
whats the slope of f on the left?, of f on the right? and of g?
anonymous
  • anonymous
slope of f on left is 1/2 and right f is -1/2? slope of g is -1?
amistre64
  • amistre64
yep
amistre64
  • amistre64
or is it 2 and -2 ..thats seems better
amistre64
  • amistre64
4 up and 2 over is 4/2 = 2
anonymous
  • anonymous
u r right did it the wrong way :)
amistre64
  • amistre64
so: g' = -1 no matter what x is f' = 2 when x<2 f' = -2 when x > 2 agreed?
anonymous
  • anonymous
correct
amistre64
  • 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
  • anonymous
so u=g(f) u'= g'(f) * f' u'= -1 * 4 = -4
amistre64
  • amistre64
when is the slope of f' = 4?
anonymous
  • anonymous
2
amistre64
  • amistre64
at x=2 ? is that your guess :) i already know why if so but its incorrect
anonymous
  • anonymous
Yes it is but I am so confused...
amistre64
  • amistre64
did we define the slope for f' at x=2?
amistre64
  • 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
  • anonymous
there is no slope there so no derivative?
amistre64
  • amistre64
correct :) no definable way to determine a derivate at a cusp, corner, pointy place
amistre64
  • amistre64
so u'2 cannot be defined either
anonymous
  • anonymous
ok let me try 3
anonymous
  • anonymous
g'(f(3))*f'(3) g'(2)(-2) (-1)(-2) 2????
amistre64
  • amistre64
exactly :)
anonymous
  • anonymous
GREAT!!! Thank you!!!
anonymous
  • anonymous
so i think I can do 56 what about 58
amistre64
  • 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
  • amistre64
D(f(g)) = f'(g) * g' define your slopes for given xs and use that to determine the answer
anonymous
  • anonymous
ok I think I can do that from the example above... it clicked in that last example... thx again for you help
amistre64
  • amistre64
your welcome :) and good luck
anonymous
  • anonymous
thx

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