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

- anonymous

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

- jamiebookeater

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- anonymous

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

- anonymous

- anonymous

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## More answers

- anonymous

anyone? anyone? bueller? bueller? lmfao

- zepdrix

lol c:

- 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

or do i differentiate with the chain rule and division rule?

- zepdrix

Yes you do that :)
Sorry I was waiting cuz I thought Canadian was going to explain it.

- anonymous

wait which one do i do? lol

- zepdrix

The second one ^^

- anonymous

Ahh I'm smarter than i give myself credit for

- anonymous

lol

- 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

minus sign of all that mess times the derivative of all that mess

- anonymous

don't forget the minus sign

- zepdrix

Woops, sorry i just did a bunch of integrals... blah :3

- 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

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

Oh he was typing after all....

- anonymous

lol go ahead zepdrix Canadian doesn't use the equation and it just hurts my eyes and confuses me.

- zepdrix

<:o

- 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

the conversion thingy ya know? I'm sorry @CanadianAsian, I thought you just switched between typing it out and the equation converter helper = (

- zepdrix

Looks like we need to apply the quotient rule to the blue part yes? :o

- anonymous

ok gotcha give me a minute let me see if I can figure it out

- anonymous

woo hoo!!~! i got it right!!

- zepdrix

Yayyy betty \c:/ boop boopy doop! or however that goes :3 ...

- anonymous

what is the last answer? i want to know..... coz i must to solve it,,,,my assignments al so the same question...

- anonymous

*also

- zepdrix

:o

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