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February 28, 2015

February 28, 2015

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

- Calculus -
**Steve**, Wednesday, October 19, 2011 at 10:36amAs 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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