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

- precal

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

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- schrodinger

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- anonymous

hint:multiply top/bottom by secx

- precal

I got
y' = (secx)/(1+secx)
but wolfram alpha does not have that as a solution

- precal

I can't multiply by secx for the top and bottom
I have to use quotient rule

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## More answers

- anonymous

sec(x) (tan^2(x)+sec^2(x)+sec(x))

- precal

No Jinnie I don't think that is correct

- anonymous

1/2 sec^2(x/2)

- anonymous

hahaha i forgot it was a division problem

- precal

ok how did you get that

- 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

y2o2 what are you using to find y prime?

- 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

-ln|cosx+1|+c ?

- precal

no iHelp that is not correct

- precal

I used a pythagorean identity for tan^2t

- 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

this is not an integration problem, I am just looking for the derivative

- anonymous

wow, srry misread the problem - my bad

- precal

its ok, you were on a roll there

- anonymous

just make sure, derivative of product is first*deri of second minus second*deri of first right?>

- anonymous

y'=sec^2x(1+secx)-tan^2xsecx , is that right ?

- precal

that is the product rule, I need to use the quotient rule.

- anonymous

nvm. miss read problem again.. wow fml...

- precal

I use lodehi-hidelo over lolo

- anonymous

last try lol... y'=(sec^2x(1+secx)-tan(x-secxtanx))/(1+secx)^2,,

- precal

yes I distributed tanx to secxtanx and got secxtan^2x
then I used the pyth. identity to get rid of tan^2x

- 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

Thanks to everyone Good night

- anonymous

=) solution page is not always right.

- anonymous

night

- 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

when do you suggest I change it to cosx? In the beginning before I take the derivative or at the end?

- 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

ok I will go back and redo the problem. Thanks for your help :)

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