Find dy/dx implicitly: x^3 - 3xy^2 + 2y = 0

[Jade]

Assume y = f(x) the equation is x^3 - 3xy^2 + 2y = 0

I came up with:

3x^2 - (6xdy/dx + 6y) + 2dy/dx = 0
dy/dx(-6x + 2) = -3x^2 + 6y
dy/dx = 3x^2 - 6y / 6x - 2

Then I am given a point (1, -1/3)
Should I plug this point into my dy/dx? :?

 

[pka]

\(\displaystyle \L x^3 - 3xy^2 + 2y = 0\quad \Rightarrow \quad 3x^2 - 3y^2 - 6xyy' + 2y' = 0\)

 

[Jade]

Thank you

You multipy any y with y'

Now you need to factor out the y' right?

Which I am coming up with y'=-3x^2+3y^2+6xy-2/2

I believe I am factoring out incorrectly.

Once I get the y', I can then plug in the point that was given to me (1, -1/3)?

 

[pka]

\(\displaystyle \L y' = \frac{{3y^2 - 3x^2 }}{{2 - 6xy}}\)

 

[Jade]

Then you could plug in a point (1,-1/3)

3(-1/3)^2-3(1)^2/2-6(1)(-1/3)

To find the equation of the tangent line after plugging in the numbers you would find slope right? Then you can plug in the slope and the point to get the equation. Am I on the right track?

y - y1 = m(x - x1)

 

[Jade]

Plugging in the point gets the slope?

When I plug in the points, I get -1/2 (is this the slope)

If I plug in the slope and the point that was given usingy-y1=m(x-x1)

would this be y-(-1/3)=-1/2(x-1)

y=-5/6(x-1)