I need help with differentiating the following equation using implicit differentiation: x^2 - 2xy + y^3 = 6
do this d/dx (x^2) d/dx(2xy) d/dx(y^3) d/dx(6)
Right, so that gives me 2x - 2y - 2y'x + 3y^2y' = 0 correct?
right. Solve for y' and you're done
This is what I get when I solve for y' but apparently its incorrect... \[y' = \frac{ 2y - 2x }{ 3y^2 - 2x }\]
You differentiated it right so it must be that you made an algebraic mistake somewhere when solving for y'. check again.
the algebra seems fine
Really? So do you think that's the correct answer?
maybe the system they are using thinks it can be reduced
Right... But I also put the equation into an online calculator and it gets ... \[y' = \frac{ 2x-2y }{ 2x - 3y^2 }\]
you can factor a 2 out of the top, but i dont see much else for it
I keep looking through my calculations and I still don't see anything wrong
factoring out a -1/-1 is not that important
2(x-y) for the top
Does it change the answer when you factor out a -1/-1 ?
it only changes form since -1/-1 = 1
(a-b) = -(b-a)
Right, so it wouldn't be a necessary step to take right?
its not required to me
x^2 - 2xy + y^3 = 6 2x x' - (2x'y+2xy') + 3y^2 y' = 0 x' = dx/dx = 1 2x - 2y -2x y' + 3y^2 y' = 0 2x - 2y + (3y^2-2x) y' = 0 2(x -y) + (3y^2-2x) y' = 0
y' = 2(y-x)/(3y^2 - 2x)
That's exactly what I did! So I don't think it should be wrong
its not wrong according to what you have provided us. can you give a screen shot or picture?
Of the question?
perferably yes ...
Sure
and how many tries do you have to get it 'correct' ?
It's unlimited
oh that good lol
Haha i know right xD
screen shot it when you input 2(y-x)/(3y^2 - 2x)
kk hold on
Oh wow now its saying that it's correct >.< This is ridiculous lol
i was going to see if you were missing a paranthesis :)
I'm always cautious of the parentheses XD there was something wrong with the program, I'm sure :D lol
Thanks so much though!
good luck :)
Thanks! :)
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