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Calculus halp

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ski93
Posts: 1279 - Karma: 83
tan(e^x)-x = cos(e^y) differentiate implicitly. Not sure if I typed this in right to alpha wolfram http://www.wolframalpha.com/input/?i=implicit+derivative+of%3A+tan%28e%5Ex%29-x+%3D+cos%28e%5Ey%29. Any help would be greatly appreciated
Apr 25 2011 10:05PM
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BootySpunk.
Posts: 8148 - Karma: 84
Too high level of calc for my knowledge.
Apr 25 2011 10:06PM
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E.C.
Posts: 3211 - Karma: 319
i forget highschool calc
Apr 25 2011 10:07PM
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purps
Posts: 468 - Karma: 30
i believe its

e^x sec^2(e^x) - 1 = -e^y sin(e^y)
Apr 25 2011 10:12PM
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ski93
Posts: 1279 - Karma: 83
this is what i got as well. The answer this sheet gives me is( -tan^2e^x)/(sine^y) hopefully it's wrong because i can't get it
Apr 25 2011 10:17PM
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Crack.
Posts: 6020 - Karma: 86
hahah AP next wednesday.. i'm so fucked
Apr 25 2011 10:18PM
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j.saaaarela
Posts: 1859 - Karma: 856
tan(e^x)-x = cos(e^y)

ok, so you would get.... e^xsec^2(e^x)-1 = e^y (dy/dx)(-sin(e^y))

and then solve for dy/dx:

(e^(x-y)sec^2(e^x)-1)/(-sin(e^y)) = dy/dx

anyone else get that? Im not sure if that's right
Apr 25 2011 10:19PM
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gregorio
Posts: 2372 - Karma: 151
same, i was in a calc class that I never did homework or studied in and still got and A. That being said most the people in my class got Cs
Apr 25 2011 10:19PM
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j.saaaarela
Posts: 1859 - Karma: 856
you need a dy/dx since it's implicit
Apr 25 2011 10:21PM
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BootySpunk.
Posts: 8148 - Karma: 84
Just kidding. It is differentiating shit from like calc 1. I completely forget how to do that. But you probably have to use trig differentiations and chain rule.
Apr 25 2011 10:21PM
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DrZoidberg
Posts: 5675 - Karma: 4,268
Yeah ^ that seems about right, but you forgot the implicit part with dy/dx and stuff, but knowing what variable you're doing with this with respect to would help. I'd guess x, you the ys need to be dy/dx right? Haven't done one of those a while.

Calc related tangent (it's like a pun or something)

I'm deep into my power series homework problems, I did the ratio test for

∞
∑ _[(-1)^n](x-1)^n
n=1 (3^2n)(ln(n))

Wow that was hard to type. But did the ratio test for that and I forgot to write down the (x-1)^n and (x-1)^(n+1) terms in the absolute value ratio thing, finished, then realized that I forgot those. fuuuuuuuuuuuuuuuuuu
Apr 25 2011 10:24PM
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purps
Posts: 468 - Karma: 30
yup yup you're very right
Apr 25 2011 11:47PM
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Sklar
Posts: 8863 - Karma: 31,577
(e^x)sec^2(e^x)-1=(e^y)dy/dx(-sin(e^y))
then solve for dy/dx


i think thats right, i could quite possibly be wrong though
Apr 25 2011 11:53PM
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Captain_Slow
Posts: 1677 - Karma: 24
ah fuck you just reminded me I got one next Thursday
Apr 26 2011 12:00AM
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DirtyDan
Posts: 1576 - Karma: 158
My brain is full of fuck!
Apr 26 2011 12:01AM
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j.saaaarela
Posts: 1859 - Karma: 856
yup that's what I got, but somehow it's supposed to be
( -tan^2e^x)/(sine^y).... I don't remember all my trig identities so I can't get any farther than solving for dy/dx with what you got.
Apr 26 2011 12:02AM
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smooticus
Posts: 1798 - Karma: 27
Ya your just not using your trig identities
Apr 26 2011 1:38AM
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