anonymous
  • anonymous
Differentiate: y=cos([1-e^2x]/[1+e^2x])
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[y=\cos(\frac{ 1-e ^{2x} }{ 1+e ^{2x} })\]
anonymous
  • anonymous
@zepdrix
anonymous
  • anonymous
@CanadianAsian

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

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