Calculus
posted by W .
The point (1,2) is on the graph of y^2x^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?

As you said, 2y(dy/dx)2x+2=0
So,
dy/dx = (2x2)/2y = (x1)/y
Noiw, just plug in the values:
y'(1) = (11)/(2) = 0
Check y''
2yy'2x+2=0
2(y')^{2} + 2yy''  2 = 0
y'' = (2  2y'^{2})/2y = (1y'^{2})/y = 1/2
Since y'' < 0 when y' = 0, it's a maximum.
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