At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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
No Jinnie I don't think that is correct
hahaha i forgot it was a division problem
ok how did you get that
y2o2 what are you using to find y prime?
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
no iHelp that is not correct
I used a pythagorean identity for tan^2t
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...
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,,
yes I distributed tanx to secxtanx and got secxtan^2x then I used the pyth. identity to get rid of tan^2x
I think I am right. I do believe wolfram alpha is not able to put all variations of a solution down. Programming issue.
Thanks to everyone Good night
=) solution page is not always right.
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
when do you suggest I change it to cosx? In the beginning before I take the derivative or at the end?
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
ok I will go back and redo the problem. Thanks for your help :)