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bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0\[y=\cos(\frac{ 1e ^{2x} }{ 1+e ^{2x} })\]

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0anyone? anyone? bueller? bueller? lmfao

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0i 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?

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0or do i differentiate with the chain rule and division rule?

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

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0wait which one do i do? lol

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0Ahh I'm smarter than i give myself credit for

zepdrix
 2 years ago
Best 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.\[\large y'=\sin \left(\frac{1e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1e^{2x}}{1+e^{2x}}\right)'}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0minus sign of all that mess times the derivative of all that mess

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0don't forget the minus sign

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Woops, sorry i just did a bunch of integrals... blah :3

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

CanadianAsian
 2 years ago
Best ResponseYou've already chosen the best response.0Differentiation 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))

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Oh he was typing after all....

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0lol go ahead zepdrix Canadian doesn't use the equation and it just hurts my eyes and confuses me.

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

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0the conversion thingy ya know? I'm sorry @CanadianAsian, I thought you just switched between typing it out and the equation converter helper = (

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Looks like we need to apply the quotient rule to the blue part yes? :o

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0ok gotcha give me a minute let me see if I can figure it out

bettyboop8904
 2 years ago
Best ResponseYou've already chosen the best response.0woo hoo!!~! i got it right!!

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.0Yayyy betty \c:/ boop boopy doop! or however that goes :3 ...

siti99
 11 months ago
Best ResponseYou've already chosen the best response.0what is the last answer? i want to know..... coz i must to solve it,,,,my assignments al so the same question...

bettyboop8904
 10 months ago
Best ResponseYou've already chosen the best response.0I 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? = )
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