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

Mathematics
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hey, how's it going?
good i guess you?
sorry, my computer crashed.

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Other answers:

its NP
Rearrange: 3xy= 5-2y^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?
it looks like you subtracted 3y from both sides but forgot to make it a negative -3y when it got to the other side
oops my bad, I guess final answer should be -3/7 then?
yep tyvm that is a choice. i have one more that i am confused on. it says "The graph of y= Ln(1-x)/(x+1) has vertical asymptotes at..." i think the answer should be at x=-1 and 1 but i dont know why.
when 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 x-value (approaches infinity)
I think you're right, at x=-1 because the denominator would = 0 and it doesn't simplify from there. At x = 1 ln(x-1)=ln(0)=undefined and approaches negative infinity as it gets closer to that value. Does that sound right?
wait no, ln(x-1) is undefined for all x <= 1, hm
lol, I messed up again, I'm too tired right now, sorry
well it was ln (1-x), right? it looks like you got it
yeah, I saw that mess up a moment ago, maybe I should just get off before I risk confusing things even more, lol
well it seems like you understand, let me know if you have any more questions
lol 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(1-x) or is there some other reason that i forgot about? sorry a cumulative test on monday so its been awhile.
well 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
well 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
in 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
ln(0) doesnt make sense either, because even e^ (-100000) is a positive number
alright so how would i know that it is ate x=1,-1 and not just at -1?
well because you know whatever is inside a natural log has to be > 0
and that for ln (x), as x gets closer to 0, ln (x) approaches -infinity, which explains the asymptote
ok 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?
sorry, 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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