Question

Differentiate:

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

Answer 1

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

Answer 2

@zepdrix

Answer 3

@CanadianAsian

Answer 4

anyone? anyone? bueller? bueller? lmfao

Answer 5

lol c:

Answer 6

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?

Answer 7

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

Answer 8

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

Answer 9

wait which one do i do? lol

Answer 10

The second one ^^

Answer 11

Ahh I'm smarter than i give myself credit for

Answer 12

lol

Answer 13

\[\large y=\cos \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\]We'll start by taking the derivative of the `outermost function`. In this case that would be the cosine part.\[\large y'=\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1-e^{2x}}{1+e^{2x}}\right)'}\]

Answer 14

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

Answer 15

don't forget the minus sign

Answer 16

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

Answer 17

\[\large y'=-\sin \left(\frac{1-e^{2x}}{1+e^{2x}}\right)\color{royalblue}{\left(\frac{1-e^{2x}}{1+e^{2x}}\right)'}\]

Answer 18

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))

Answer 19

Oh he was typing after all....

Answer 20

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

Answer 21

<:o

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

*also

Answer 30

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? = )

Answer 31

:o