Question

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

Answer 1

hint:multiply top/bottom by secx

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

No Jinnie I don't think that is correct

Answer 6

1/2 sec^2(x/2)

Answer 7

hahaha i forgot it was a division problem

Answer 8

ok how did you get that

Answer 9

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

Answer 10

y2o2 what are you using to find y prime?

Answer 11

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

Answer 12

-ln|cosx+1|+c ?

Answer 13

no iHelp that is not correct

Answer 14

I used a pythagorean identity for tan^2t

Answer 15

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...

Answer 16

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

Answer 17

wow, srry misread the problem - my bad

Answer 18

its ok, you were on a roll there

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

I use lodehi-hidelo over lolo

Answer 24

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

Answer 25

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

Answer 26

I think I am right. I do believe wolfram alpha is not able to put all variations of a solution down. Programming issue.

Answer 27

Thanks to everyone Good night

Answer 28

=) solution page is not always right.

Answer 29

night

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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