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hint:multiply top/bottom by secx

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

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

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

No Jinnie I don't think that is correct

1/2 sec^2(x/2)

hahaha i forgot it was a division problem

ok how did you get that

\[{-\sec(x) \tan(x)^2\over(\sec(x)+1)^2}+{\sin(x)^2\over(\cos(x)^2}+{\cos(x))+1\over(\sec(x)+1)}\]

y2o2 what are you using to find y prime?

-ln|cosx+1|+c ?

no iHelp that is not correct

I used a pythagorean identity for tan^2t

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

wow, srry misread the problem - my bad

its ok, you were on a roll there

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

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

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

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

I use lodehi-hidelo over lolo

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

Thanks to everyone Good night

=) solution page is not always right.

night

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

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