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I need someone to explain to me step by step the following problem:

Mathematics
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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 have the problem worked out in front of me (prof gave us the answers) but, his work is disorganized and I don't understand what it is he's doing.
I figure I have to do this implicitly.

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Other answers:

yes - you are correct - use implicit differentiation to get an expression for slope of the tangent line at any point (x,y). then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1) then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)
is the answer y=4-5x?
that's not the answer.... the answer is: y=-1/5x+4/5
its a plane....?
okay and the way my prof did his differentiation was really weird. his first step looks like: x^{2}2y*y'+y^{2}(2x)=3(1+y')
a plane? huh?
that's not how i would start the implicit differentiation.
\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]
what your prof did is correct - he is using the product rule, see here: http://www.1728.org/chainrul.htm
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 you can effectively just ignore the 3 outside the braces and just differentiate what is inside the braces
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?
yes - what you need to do is to get an expression for y'. so move all terms involving y' to the left and the rest to the right
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
you started with this:\[x^{2}(2y*y')+y^{2}(2x)=3(1+y')\]so first expand the right to get:\[x^{2}(2y*y')+y^{2}(2x)=3+3y'\]then move terms involving y' to the left and the rest to the right to get:\[x^{2}(2y*y')-3y'=3-y^{2}(2x)\]
then factorise y' on the left
\[y'=\frac{ 3-2xy^{2} }{ 2x^{2}y-3 }\]
perfect!
woooo. so now that part is done, right?
yup - the rest should be straight forward. I need to go for dinner now. I'll check back afterwards to see if you had any further problems.
the steps left are: then substitute x=1, y=-1 to find the value of the slope of the tangent line at (1, -1) then use the equation of a straight line: y = mx + c to find c (since you now know m) by using the fact that it passes through the point (1, -1)
okay thanks. : )
yw :) and good luck!
i got the rest of it now ; ) thanks a bunch!
thats great! and you are more than welcome! :)

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