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Differentiate: y=cos([1-e^2x]/[1+e^2x])

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\[y=\cos(\frac{ 1-e ^{2x} }{ 1+e ^{2x} })\]

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Other answers:

anyone? anyone? bueller? bueller? lmfao
lol c:
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?
or do i differentiate with the chain rule and division rule?
Yes you do that :) Sorry I was waiting cuz I thought Canadian was going to explain it.
wait which one do i do? lol
The second one ^^
Ahh I'm smarter than i give myself credit for
\[\large y=\cos \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\]We'll start by taking the derivative of the `outermost function`. In this case that would be the cosine part.\[\large y'=\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1-e^{2x}}{1+e^{2x}}\right)'}\]
minus sign of all that mess times the derivative of all that mess
don't forget the minus sign
Woops, sorry i just did a bunch of integrals... blah :3
\[\large y'=-\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1-e^{2x}}{1+e^{2x}}\right)'}\]
Differentiation rules tell us that we have to work our way in. So for the first part let a = \[(1-e^(2x))/(1+ e^(2x))\] Take the derivative of the cos(a) Which gives us -sin(a) Plug the actual value of a back in to get \[-sin((1-e^(2x))/(1+ e^(2x))\] 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)*(1-e^(2x)))/(1+e^(2x))^2 Multiply that by -sin((1-e^2x)/(1+e^2x))
Oh he was typing after all....
lol go ahead zepdrix Canadian doesn't use the equation and it just hurts my eyes and confuses me.
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. \[\large y'=-\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1-e^{2x}}{1+e^{2x}}\right)'}\]
the conversion thingy ya know? I'm sorry @CanadianAsian, I thought you just switched between typing it out and the equation converter helper = (
Looks like we need to apply the quotient rule to the blue part yes? :o
ok gotcha give me a minute let me see if I can figure it out
woo hoo!!~! i got it right!!
Yayyy betty \c:/ boop boopy doop! or however that goes :3 ...
what is the last answer? i want to know..... coz i must to solve it,,,,my assignments al so the same question...
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? = )

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