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implicit differentiation help. Somebody here has got to know how to do this problem... >.< http://gyazo.com/53dfb49c3ca8d1a3b949b4f9f2afa67c

Mathematics
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Do you know the power rule?
yes
i know how to get y' alone. im having problems putting it in the formular y-y1=m(x-x1)

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Other answers:

How so?
(2/3) * x^-(1/3) + (2/3) * y^-(1/3) * (dy/dx) = 0 dy/dx = -x^(-1/3) / y^(-1/3)
yes, got that far. now what coolsector?
Fill in y
and x
plug x = -3sqrt(3) , y= 1
i filled in both x and y. and got a number. i used that as my slope in the equation y-y1=m(x-x1)
you get dy/dx = - (3sqrt(3))^(-1/3)
yes, which is -1.732 corect?
mm i ges -0.577 in fact
and x1 = -3sqrt(3) y1 = 1...
dy/dx = -x^(-1/3) / y^(-1/3) i replaced x and y with given points in this, and got -1.732
dy/dx = -(3sqrt(3))^(-1/3) / 1
dy/dx = -(3sqrt(3))^(-1/3) i get -0.577
ok. go on with that then. what is the final answer?
need an equation in y= form
y =1 -0.577(x+3sqrt(3)) y= 1 -0.577x -3 y = -0.577x -2
in fact dy/dx = -1/sqrt(3) so : y = -x/sqrt(3) -2
\[y = \frac{ -x }{ \sqrt{3} } -2\]
http://gyazo.com/b7dc43e97670c3b7879e2ed74d7c4fe0
says incorrect :/
if we plug x = -3sqrt(3) we get y = 1 right ?
yes
well weird
yeah, i have no idea how to do this problem :/ thanks anyways
im sure it is correct
just tried putting in again, says wrong. :|
going to close post and make another one.

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