Calculus
posted by Karen .
Find dy/dx for: 2y^3  3xy = 4
Write an equasion for the line tangent to the curve at (1,2)
Find d^2y/dx^2 at (1,2)

As you can see, it is not easy to solve for y in terms of x only and then differentiate that equation. It is much easier to use the method called "implicit differentiation" in which you take the derivative of both sides of the equation, treating y as a function of x. This results in:
6y^2*dy/dx  3x*dy/dx 3y = 0
dy/dx*(2y^2x) = y
dy/dx = y/(2y^2x)
At (1,2), dy/dx = 2/(81) = 2/7
Use that slope and the coordinates (1,2) that the line must pass through to get the equation of the tangent line.
For the second derivative, differentiate an equation containing dy/dx impliticly with respect to x.
dy/dx*(2y^2x) = y
d^2y/dx^2*(2y^2x)
+ dy/dx (4y*dy/dx 1) = dy/dx
Insert the value of dy/dx = 2/7 that you already know at the point (1,2), and solve for d^2y/dx^2 at the same point. 
I did that and got 2... my friend said the answer is just 2/7. Is she right?