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

- anonymous

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

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- anonymous

hey, how's it going?

- anonymous

good i guess you?

- anonymous

sorry, my computer crashed.

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## More answers

- anonymous

its NP

- anonymous

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?

- anonymous

it looks like you subtracted 3y from both sides but forgot to make it a negative -3y when it got to the other side

- anonymous

oops my bad, I guess final answer should be -3/7 then?

- anonymous

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.

- anonymous

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)

- anonymous

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?

- anonymous

wait no, ln(x-1) is undefined for all x <= 1, hm

- anonymous

lol, I messed up again, I'm too tired right now, sorry

- anonymous

well it was ln (1-x), right? it looks like you got it

- anonymous

yeah, I saw that mess up a moment ago, maybe I should just get off before I risk confusing things even more, lol

- anonymous

well it seems like you understand, let me know if you have any more questions

- anonymous

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.

- anonymous

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

- anonymous

- anonymous

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

- anonymous

ln(0) doesnt make sense either, because even e^ (-100000) is a positive number

- anonymous

alright so how would i know that it is ate x=1,-1 and not just at -1?

- anonymous

well because you know whatever is inside a natural log has to be > 0

- anonymous

and that for ln (x), as x gets closer to 0, ln (x) approaches -infinity, which explains the asymptote

- anonymous

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?

- anonymous

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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