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precal

  • 4 years ago

what is the derivative of y=(tanx)/(1+secx)

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  1. anonymous
    • 4 years ago
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    hint:multiply top/bottom by secx

  2. precal
    • 4 years ago
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    I got y' = (secx)/(1+secx) but wolfram alpha does not have that as a solution

  3. precal
    • 4 years ago
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    I can't multiply by secx for the top and bottom I have to use quotient rule

  4. anonymous
    • 4 years ago
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    sec(x) (tan^2(x)+sec^2(x)+sec(x))

  5. precal
    • 4 years ago
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    No Jinnie I don't think that is correct

  6. anonymous
    • 4 years ago
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    1/2 sec^2(x/2)

  7. anonymous
    • 4 years ago
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    hahaha i forgot it was a division problem

  8. precal
    • 4 years ago
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    ok how did you get that

  9. y2o2
    • 4 years ago
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    \[{-\sec(x) \tan(x)^2\over(\sec(x)+1)^2}+{\sin(x)^2\over(\cos(x)^2}+{\cos(x))+1\over(\sec(x)+1)}\]

  10. precal
    • 4 years ago
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    y2o2 what are you using to find y prime?

  11. anonymous
    • 4 years ago
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    it does work at walfarm, check this out: http://www.wolframalpha.com/input/?i=d%2F+dx+%28sinx%2Fcosx%29%2F%281%2Bsecx%29&cdf=1

  12. anonymous
    • 4 years ago
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    -ln|cosx+1|+c ?

  13. precal
    • 4 years ago
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    no iHelp that is not correct

  14. precal
    • 4 years ago
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    I used a pythagorean identity for tan^2t

  15. anonymous
    • 4 years ago
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    tanx/(secx+1)=sinx/cosx+1, let u=cosx+1, du=-sinxdx, so integral of -1/u, which is -ln|u| , subisitute u back in...

  16. precal
    • 4 years ago
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    this is not an integration problem, I am just looking for the derivative

  17. anonymous
    • 4 years ago
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    wow, srry misread the problem - my bad

  18. precal
    • 4 years ago
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    its ok, you were on a roll there

  19. anonymous
    • 4 years ago
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    just make sure, derivative of product is first*deri of second minus second*deri of first right?>

  20. anonymous
    • 4 years ago
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    y'=sec^2x(1+secx)-tan^2xsecx , is that right ?

  21. precal
    • 4 years ago
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    that is the product rule, I need to use the quotient rule.

  22. anonymous
    • 4 years ago
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    nvm. miss read problem again.. wow fml...

  23. precal
    • 4 years ago
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    I use lodehi-hidelo over lolo

  24. anonymous
    • 4 years ago
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    last try lol... y'=(sec^2x(1+secx)-tan(x-secxtanx))/(1+secx)^2,,

  25. precal
    • 4 years ago
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    yes I distributed tanx to secxtanx and got secxtan^2x then I used the pyth. identity to get rid of tan^2x

  26. precal
    • 4 years ago
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    I think I am right. I do believe wolfram alpha is not able to put all variations of a solution down. Programming issue.

  27. precal
    • 4 years ago
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    Thanks to everyone Good night

  28. anonymous
    • 4 years ago
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    =) solution page is not always right.

  29. anonymous
    • 4 years ago
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    night

  30. dumbcow
    • 4 years ago
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    yes you are correct...i also get dy/dx = sec/(1+sec) you can change it in terms of cos --> dy/dx = 1/(1+cos) and this is one of the forms given on wolframalpha

  31. precal
    • 4 years ago
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    when do you suggest I change it to cosx? In the beginning before I take the derivative or at the end?

  32. dumbcow
    • 4 years ago
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    it doesn't really matter...it depends on the function, in this case if you changed it before it would have made taking the derivative simpler

  33. precal
    • 4 years ago
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    ok I will go back and redo the problem. Thanks for your help :)

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