Post a New Question

Calculus

posted by on .

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?

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question