ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 that's the only reason I can think of where the negative comes from Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 i forgot that derivative of 1/x is -1/x^2.thanks for all your help... Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 trust me, it can be confusing at times Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 yes its true. its confusing sometimes... you thought that your solutions are correct but when you review it carefully, there's the errors. its happening to me often. thats why i've got wrong answers. Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 you'll get used to this soon enough Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 i always commit mistakes. i want it to remove in me. Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 if you have more feel free to post them here, I'll see what I can do Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 let's see this problem... ∫ dx / (1+e^x) i've got the correct answer which is -ln (1 + e^-x) with solutions unsure... Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 same concept: think dx/(1+e^x) as (1+e^x)^-1 dx Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 but u = 1 + e^xand du = e^xthere is no e^x in the expression Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 that's because you don't use substitution Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 what i do is e^x can be 1/ e^-x and then simplifythen will become e^-x dx / (e^-x +1) where u = e^-x + 1 and du = -e^-x there is the du.can it be? do i made an illegal act? Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 technically you can do that, though I think you just made it harder than the problem really is Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 its because that's the only way i know to solve that... u substitution is only i knowhow about if it will be solved by part? Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 that wouldn't matter since the integration and derivative of e^x is still e^x Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 thanks again... can you solve physics problem? i only ask Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 you put it here and let me take a look Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 i have not yet... i only ask... Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 Then if you get stuck I'll try it Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 yes... thanks... that's all... are you a genius? Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 of course not, I just happened to know what you're talking about Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 are you a student? Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 obviously, and that's why I know what you're talking about Link to comment Share on other sites More sharing options...
gutswarrior Posted November 23, 2008 Author Report Share Posted November 23, 2008 what year are you now and what course? Link to comment Share on other sites More sharing options...
ragnarok1945 Posted November 23, 2008 Report Share Posted November 23, 2008 that's off the topic and not for you to know. Anything else you want to put here? Link to comment Share on other sites More sharing options...
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