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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)

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.