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Find the equation of the tangent line to the curve
\[x^{2}y^{2}-1=3(x+y)\] at the points (1,-1)

I figure I have to do this implicitly.

is the answer y=4-5x?

that's not the answer.... the answer is: y=-1/5x+4/5

its a plane....?

a plane? huh?

that's not how i would start the implicit differentiation.

\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]

how would you have done this?

haha no idea... i suck at math : (

so i use the product rule for x^2*y^2 ??

correct

and for the 3*(x+y) ??

that is just straight differentiation since it can be written as: 3x + 3y

so that will just equal 3y' ??

or 3+3y'

so it's 3(1+y')

yes - correct

okay so now i have the same first line as he does haha.

i ignored the 3 outside the brackets because it's a constant right?

yes.
thats good - you are making progress - maybe you are a lot better at maths than you think. :)

haha maybe, maybe not...

so now i can take the y^2(2x) over to the other side?

and the right hand side over to the left?

first expand the expression on the right

do i leave my 3 on the right hand side?

snap! yes :)

okay so now i have: (x^2)(2y*y')-(1-y')=3(-2xy^2)

that look right to you?

not quite :(

oh snappp

then factorise y' on the left

\[y'=\frac{ 3-2xy^{2} }{ 2x^{2}y-3 }\]

perfect!

woooo. so now that part is done, right?

okay thanks. : )

yw :) and good luck!

i got the rest of it now ; ) thanks a bunch!

thats great!
and you are more than welcome! :)