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anonymous
 5 years ago
If 3xy+2y^2=5, find (DY/DX) at (1,1)
anonymous
 5 years ago
If 3xy+2y^2=5, find (DY/DX) at (1,1)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, my computer crashed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Rearrange: 3xy= 52y^2 Differentiate implicitly: 3y+3xdy/dx=4ydy/dx Rearrange again: (3x+4y)dy/dx=3y Solve for dy/dx dy/dx=3y/(3x+4y) for the point (1,1) substitute point for x and y: dy/dx=3(1)/[3(1)+4(1)]=3/7 does this make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it looks like you subtracted 3y from both sides but forgot to make it a negative 3y when it got to the other side

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops my bad, I guess final answer should be 3/7 then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep tyvm that is a choice. i have one more that i am confused on. it says "The graph of y= Ln(1x)/(x+1) has vertical asymptotes at..." i think the answer should be at x=1 and 1 but i dont know why.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when they ask for asymptotes, you want to find where the function is undefined. the first thing to check is setting the denominator of a fraction = to 0, since the function is undefined at that xvalue (approaches infinity)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you're right, at x=1 because the denominator would = 0 and it doesn't simplify from there. At x = 1 ln(x1)=ln(0)=undefined and approaches negative infinity as it gets closer to that value. Does that sound right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait no, ln(x1) is undefined for all x <= 1, hm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol, I messed up again, I'm too tired right now, sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well it was ln (1x), right? it looks like you got it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, I saw that mess up a moment ago, maybe I should just get off before I risk confusing things even more, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well it seems like you understand, let me know if you have any more questions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol it alright i guess the only thing that is confusing me on this is if your looking for the vertical why would you look in the numerator? because it is Ln(1x) or is there some other reason that i forgot about? sorry a cumulative test on monday so its been awhile.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well because you have the natural log function, you know that can't end up negative after you plug in x, because the natural log means that e^y = whatever is inside the natural log, which will always be a positive number since e is positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well because you have the natural log function, you know that can't end up negative after you plug in x, because the natural log means that e^y = whatever is inside the natural log, which will always be a positive number since e is positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in other words, ln of a negative number doesn't make sense, since there is no exponent that you could raise e to and make it come out negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ln(0) doesnt make sense either, because even e^ (100000) is a positive number

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright so how would i know that it is ate x=1,1 and not just at 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well because you know whatever is inside a natural log has to be > 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and that for ln (x), as x gets closer to 0, ln (x) approaches infinity, which explains the asymptote

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so because i know that there is a vertical asymptote at x=1 to make the bottom =0 but i also need to make the top a 0 becasue it is a Ln function and i know it needs to be 0 because ln(0) is also a Vert asymptote?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, my computer crashed again. you basically have it right, except "ln(0) is a vert asymptote" isn't quite right, more like f(x) = ln x has a vert asymptote at x = 0
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