# Calculus

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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)

• Calculus - ,

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^2-x) = y
dy/dx = y/(2y^2-x)
At (1,2), dy/dx = 2/(8-1) = 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^2-x) = y
d^2y/dx^2*(2y^2-x)
+ 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.

• Calculus - ,

I did that and got 2... my friend said the answer is just 2/7. Is she right?