precal
  • precal
what is the derivative of y=(tanx)/(1+secx)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
hint:multiply top/bottom by secx
precal
  • precal
I got y' = (secx)/(1+secx) but wolfram alpha does not have that as a solution
precal
  • precal
I can't multiply by secx for the top and bottom I have to use quotient rule

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anonymous
  • anonymous
sec(x) (tan^2(x)+sec^2(x)+sec(x))
precal
  • precal
No Jinnie I don't think that is correct
anonymous
  • anonymous
1/2 sec^2(x/2)
anonymous
  • anonymous
hahaha i forgot it was a division problem
precal
  • precal
ok how did you get that
y2o2
  • y2o2
\[{-\sec(x) \tan(x)^2\over(\sec(x)+1)^2}+{\sin(x)^2\over(\cos(x)^2}+{\cos(x))+1\over(\sec(x)+1)}\]
precal
  • precal
y2o2 what are you using to find y prime?
anonymous
  • anonymous
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
anonymous
  • anonymous
-ln|cosx+1|+c ?
precal
  • precal
no iHelp that is not correct
precal
  • precal
I used a pythagorean identity for tan^2t
anonymous
  • anonymous
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...
precal
  • precal
this is not an integration problem, I am just looking for the derivative
anonymous
  • anonymous
wow, srry misread the problem - my bad
precal
  • precal
its ok, you were on a roll there
anonymous
  • anonymous
just make sure, derivative of product is first*deri of second minus second*deri of first right?>
anonymous
  • anonymous
y'=sec^2x(1+secx)-tan^2xsecx , is that right ?
precal
  • precal
that is the product rule, I need to use the quotient rule.
anonymous
  • anonymous
nvm. miss read problem again.. wow fml...
precal
  • precal
I use lodehi-hidelo over lolo
anonymous
  • anonymous
last try lol... y'=(sec^2x(1+secx)-tan(x-secxtanx))/(1+secx)^2,,
precal
  • precal
yes I distributed tanx to secxtanx and got secxtan^2x then I used the pyth. identity to get rid of tan^2x
precal
  • precal
I think I am right. I do believe wolfram alpha is not able to put all variations of a solution down. Programming issue.
precal
  • precal
Thanks to everyone Good night
anonymous
  • anonymous
=) solution page is not always right.
anonymous
  • anonymous
night
dumbcow
  • dumbcow
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
precal
  • precal
when do you suggest I change it to cosx? In the beginning before I take the derivative or at the end?
dumbcow
  • dumbcow
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
precal
  • precal
ok I will go back and redo the problem. Thanks for your help :)

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