Posted by **W** on Tuesday, October 18, 2011 at 10:27pm.

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 -
**Steve**, Wednesday, October 19, 2011 at 10:36am
As you said, 2y(dy/dx)-2x+2=0

So,

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

Noiw, just plug in the values:

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

Check y''

2yy'-2x+2=0

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.

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