Post a New Question


posted by .

The point (1,-2) is on the graph of y^2-x^2+2x=5. Find the value of (dy/dx) and (d^2y/dx^2) at the point (1,-2). How do I find (d^2y/dx^2)? I found. I found (dy/dx) to be 2y(dy/dx)-2x+2=0, but I don't know how to take the derivative of that to get(d^2y/dx^2)? I'm not sure if that is correct, but that's just my thinking. Also, how do i find if the graph has a relative min or max?

  • Calculus -

    As you said, 2y(dy/dx)-2x+2=0


    dy/dx = (2x-2)/2y = (x-1)/y

    Noiw, just plug in the values:

    y'(1) = (1-1)/(-2) = 0

    Check y''


    2(y')2 + 2yy'' - 2 = 0
    y'' = (2 - 2y'2)/2y = (1-y'2)/y = 1/-2

    Since y'' < 0 when y' = 0, it's a maximum.

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question