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bettyboop8904Best ResponseYou've already chosen the best response.0
\[y=\cos(\frac{ 1e ^{2x} }{ 1+e ^{2x} })\]
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
anyone? anyone? bueller? bueller? lmfao
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
i was thinking looking at it like a horizontal asymptote? but isn't that only when there's a limit in front of it? where you would divide everything by the highest exponent which is e^2x so you would be left with 1/1 which is cos(1) is that right?
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
or do i differentiate with the chain rule and division rule?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Yes you do that :) Sorry I was waiting cuz I thought Canadian was going to explain it.
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
wait which one do i do? lol
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
Ahh I'm smarter than i give myself credit for
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large y=\cos \left(\frac{1e^{2x}}{1+e^{2x}}\right)\]We'll start by taking the derivative of the `outermost function`. In this case that would be the cosine part.\[\large y'=\sin \left(\frac{1e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1e^{2x}}{1+e^{2x}}\right)'}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
minus sign of all that mess times the derivative of all that mess
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
don't forget the minus sign
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Woops, sorry i just did a bunch of integrals... blah :3
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large y'=\sin \left(\frac{1e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1e^{2x}}{1+e^{2x}}\right)'}\]
 one year ago

CanadianAsianBest ResponseYou've already chosen the best response.0
Differentiation rules tell us that we have to work our way in. So for the first part let a = \[(1e^(2x))/(1+ e^(2x))\] Take the derivative of the cos(a) Which gives us sin(a) Plug the actual value of a back in to get \[sin((1e^(2x))/(1+ e^(2x))\] Now let's do the inside. The derivative of e^2x is 2e^2x. And the derivative of f(x)/g(x) is: (f'(x)g(x)  g'(x)f(x))/g(x)^2 So this gives us (2e^(2x)*(1+e^(2x)) 2e^(2x)*(1e^(2x)))/(1+e^(2x))^2 Multiply that by sin((1e^2x)/(1+e^2x))
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Oh he was typing after all....
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
lol go ahead zepdrix Canadian doesn't use the equation and it just hurts my eyes and confuses me.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
So like Canadian was saying, we work our way from the outside in c: Taking the derivative of the `outermost function` gave us the sine. Then, according to the `chain rule` we have to make a copy of the inside, and multiply by it's derivative. That's what the blue part represents. The little prime on the outside is to let us know that we still need to differentiate it. \[\large y'=\sin \left(\frac{1e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1e^{2x}}{1+e^{2x}}\right)'}\]
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
the conversion thingy ya know? I'm sorry @CanadianAsian, I thought you just switched between typing it out and the equation converter helper = (
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Looks like we need to apply the quotient rule to the blue part yes? :o
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
ok gotcha give me a minute let me see if I can figure it out
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
woo hoo!!~! i got it right!!
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Yayyy betty \c:/ boop boopy doop! or however that goes :3 ...
 one year ago

siti99Best ResponseYou've already chosen the best response.0
what is the last answer? i want to know..... coz i must to solve it,,,,my assignments al so the same question...
 one month ago

bettyboop8904Best ResponseYou've already chosen the best response.0
I haven't done math since that class a year ago and sadly I forget how to do it. @zepdrix can you help @siti99 with this problem? = )
 one month ago
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