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bettyboop8904

  • 2 years ago

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

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  1. bettyboop8904
    • 2 years ago
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    \[y=\cos(\frac{ 1-e ^{2x} }{ 1+e ^{2x} })\]

  2. bettyboop8904
    • 2 years ago
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    @zepdrix

  3. bettyboop8904
    • 2 years ago
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    @CanadianAsian

  4. bettyboop8904
    • 2 years ago
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    anyone? anyone? bueller? bueller? lmfao

  5. zepdrix
    • 2 years ago
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    lol c:

  6. bettyboop8904
    • 2 years ago
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    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?

  7. bettyboop8904
    • 2 years ago
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    or do i differentiate with the chain rule and division rule?

  8. zepdrix
    • 2 years ago
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    Yes you do that :) Sorry I was waiting cuz I thought Canadian was going to explain it.

  9. bettyboop8904
    • 2 years ago
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    wait which one do i do? lol

  10. zepdrix
    • 2 years ago
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    The second one ^^

  11. bettyboop8904
    • 2 years ago
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    Ahh I'm smarter than i give myself credit for

  12. bettyboop8904
    • 2 years ago
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    lol

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

  14. satellite73
    • 2 years ago
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    minus sign of all that mess times the derivative of all that mess

  15. satellite73
    • 2 years ago
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    don't forget the minus sign

  16. zepdrix
    • 2 years ago
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    Woops, sorry i just did a bunch of integrals... blah :3

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

  18. CanadianAsian
    • 2 years ago
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    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))

  19. zepdrix
    • 2 years ago
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    Oh he was typing after all....

  20. bettyboop8904
    • 2 years ago
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    lol go ahead zepdrix Canadian doesn't use the equation and it just hurts my eyes and confuses me.

  21. zepdrix
    • 2 years ago
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    <:o

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

  23. bettyboop8904
    • 2 years ago
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    the conversion thingy ya know? I'm sorry @CanadianAsian, I thought you just switched between typing it out and the equation converter helper = (

  24. zepdrix
    • 2 years ago
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    Looks like we need to apply the quotient rule to the blue part yes? :o

  25. bettyboop8904
    • 2 years ago
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    ok gotcha give me a minute let me see if I can figure it out

  26. bettyboop8904
    • 2 years ago
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    woo hoo!!~! i got it right!!

  27. zepdrix
    • 2 years ago
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    Yayyy betty \c:/ boop boopy doop! or however that goes :3 ...

  28. siti99
    • one year ago
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    what is the last answer? i want to know..... coz i must to solve it,,,,my assignments al so the same question...

  29. siti99
    • one year ago
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    *also

  30. bettyboop8904
    • one year ago
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    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? = )

  31. zepdrix
    • one year ago
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    :o

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