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It's been a year since I touched math and my forgetfulness of even the most basic calc knowledge is beyond frustrating.
I'm re-learning most of it all over again, and so far I've been stuck on 3 questions for my assignment:
Find the second derivative of each of the following functions:
a. f(x) = e^(e^x)
b. g(y) = y^y for y > 0
c. h(t) = ln[t^(1/2)] for t > 0
My fail solutions so far:
a. f'(x)=e^(e^x) * (e^x)
f’’(x)=e^(e^x) * e^x * e^x + e^(e^x) * e^x
= can I simply further?
b. g(y)= y^y= e ^ (y * ln(y))
g’(y)= e ^ (y * ln(y)) * d/dx(y * ln(y))
= e ^ (y * ln(y)) * (ln(y) + y/y))
= e ^ (y * ln(y)) * (ln(y) + 1)
= ? (How do I simply and get the second derivative?)
c. Not a clue.
If someone could correct my answers and point me in the right direction it would be greatly appreciated, I’ve been working at this for 6 hours straight already and I’m going crazy:S
Help me please the brains of TL 
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i could be wrong, because its been a while, but i'm pretty sure homework blogs are generally discouraged.
when i enter into TI-89
a) infinity
b) y^(y-1) * (y * (ln(y))^2 + 2*y*ln(y) + y + 1)
c) 0
EDIT: yeah, Commandment #5
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a) Apart from messing with indices a little that seems to be as good as you'll get.
b) Well you already know what the ( +1) bit looks like, just use the product rule on the other part. It's just like (a) Alternatively you can use the product rule with (e^ylny) and (lny+1). i.e. not expanding the bracket.
c) Just use the chain rule like you have been doing before.
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On June 07 2010 07:55 majesty.k)seRapH wrote:i could be wrong, because its been a while, but i'm pretty sure homework blogs are generally discouraged. when i enter into TI-89 a) infinity b) y^(y-1) * (y * (ln(y))^2 + 2*y*ln(y) + y + 1) c) 0 EDIT: yeah, Commandment #5
Yeah homework threads are frowned upon, but a good solution to that is to check out the Teamliquid Manpower thread, there are plenty of people you can PM for calculus help (I'm one of them ).
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You already know everything you need, you just don't realize it! This won't be as hard as you're making it out to be
a) You can 'simplify' further by adding the exponents. a^b * a^c = a^(b+c) , if you like how that looks better. You have the right answer, though.
b) At the end you just have g' = g * (ln(y) + 1). You know the product rule, you know the derivative of g, so just go for it.
c) I see you know the derivative of ln(_) from part b). And I see you know the chain rule from part a). Just combine the two. Let t^(1/2) = u.
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Do as ninjafetus says, if you're completely lost, I did c for you, check in spoilers if you cannot figure it out:
+ Show Spoiler +c) h'(t)=-1/2*t^(-1/2)*1/(t^1/2)=-1/2*1/t, because of chainrule
h''(t)= 1/4*1/t^2
This is just a simple use of the fact that (f(g(x)))'=g'(x)f'(g(x)) aka the chainrule
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Nytefish is correct in his clues.
I think most homework threads are frowned on but somewhat tolerated if its not so blatant that you're not willing to work on it. I think that showing your work and getting some people to point you in the right direction like above at least imo is ok.
Yeah I would shy away from actually doing the problem like Papvin does above, albeit in spoilers... its not helping the person learn what they need to.
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On June 07 2010 08:07 Papvin wrote:Do as ninjafetus says, if you're completely lost, I did c for you, check in spoilers if you cannot figure it out: + Show Spoiler +c) h'(t)=-1/2*t^(-1/2)*1/(t^1/2)=-1/2*1/t, because of chainrule
h''(t)= 1/4*1/t^2
This is just a simple use of the fact that (f(g(x)))'=g'(x)f'(g(x)) aka the chainrule
+ Show Spoiler +btw there's an extra minus sign in this
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YES!
Thank you guys.
All of you, are my personal Jesus Christs, true saviors.
<3 TL. I'll make sure to give back to the community in one form or another
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EPT
