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bettyboop8904 Group Title

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

  • one year ago
  • one year ago

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

    • one year ago
  2. bettyboop8904 Group Title
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    @zepdrix

    • one year ago
  3. bettyboop8904 Group Title
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    @CanadianAsian

    • one year ago
  4. bettyboop8904 Group Title
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    anyone? anyone? bueller? bueller? lmfao

    • one year ago
  5. zepdrix Group Title
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    lol c:

    • one year ago
  6. bettyboop8904 Group Title
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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?

    • one year ago
  7. bettyboop8904 Group Title
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    or do i differentiate with the chain rule and division rule?

    • one year ago
  8. zepdrix Group Title
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    Yes you do that :) Sorry I was waiting cuz I thought Canadian was going to explain it.

    • one year ago
  9. bettyboop8904 Group Title
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    wait which one do i do? lol

    • one year ago
  10. zepdrix Group Title
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    The second one ^^

    • one year ago
  11. bettyboop8904 Group Title
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    Ahh I'm smarter than i give myself credit for

    • one year ago
  12. bettyboop8904 Group Title
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    lol

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

    • one year ago
  14. satellite73 Group Title
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    minus sign of all that mess times the derivative of all that mess

    • one year ago
  15. satellite73 Group Title
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    don't forget the minus sign

    • one year ago
  16. zepdrix Group Title
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    Woops, sorry i just did a bunch of integrals... blah :3

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

    • one year ago
  18. CanadianAsian Group Title
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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))

    • one year ago
  19. zepdrix Group Title
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    Oh he was typing after all....

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

    • one year ago
  21. zepdrix Group Title
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    <:o

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

    • one year ago
  23. bettyboop8904 Group Title
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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 = (

    • one year ago
  24. zepdrix Group Title
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    Looks like we need to apply the quotient rule to the blue part yes? :o

    • one year ago
  25. bettyboop8904 Group Title
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    ok gotcha give me a minute let me see if I can figure it out

    • one year ago
  26. bettyboop8904 Group Title
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    woo hoo!!~! i got it right!!

    • one year ago
  27. zepdrix Group Title
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    Yayyy betty \c:/ boop boopy doop! or however that goes :3 ...

    • one year ago
  28. siti99 Group Title
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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...

    • 6 months ago
  29. siti99 Group Title
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    *also

    • 6 months ago
  30. bettyboop8904 Group Title
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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? = )

    • 5 months ago
  31. zepdrix Group Title
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    :o

    • 5 months ago
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