## precal 4 years ago what is the derivative of y=(tanx)/(1+secx)

1. anonymous

hint:multiply top/bottom by secx

2. precal

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

3. precal

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

4. anonymous

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

5. precal

No Jinnie I don't think that is correct

6. anonymous

1/2 sec^2(x/2)

7. anonymous

hahaha i forgot it was a division problem

8. precal

ok how did you get that

9. y2o2

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

10. precal

y2o2 what are you using to find y prime?

11. 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

12. anonymous

-ln|cosx+1|+c ?

13. precal

no iHelp that is not correct

14. precal

I used a pythagorean identity for tan^2t

15. 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...

16. precal

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

17. anonymous

18. precal

its ok, you were on a roll there

19. anonymous

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

20. anonymous

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

21. precal

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

22. anonymous

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

23. precal

I use lodehi-hidelo over lolo

24. anonymous

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

25. precal

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

26. precal

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

27. precal

Thanks to everyone Good night

28. anonymous

=) solution page is not always right.

29. anonymous

night

30. anonymous

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

31. precal

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

32. anonymous

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

33. precal

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