Jawabanmu

2014-01-06T22:38:26+07:00
(x-y)^2-y=3\\
x^2-2xy+y^2-y=3\\
 \frac{d}{dx}( x^2-2xy+y^2-y)= \frac{d}{dx}( 3)\\
2x-[(2x)'(y)+(2x)(y)']+2y.y'-1.y'=0\\
2x-(2y+2xy')+2y.y'-y'=0\\
2x-2y-2xy'+2y.y'-y'=0\\
(-2x+2y-1)y'+2x-2y=0\\
(-2x+2y-1)y'=-2x+2y\\
 y' = \frac{-2x+2y}{-2x+2y-1} \\\\
y'= \frac{2x-2y}{2x-2y+1} \\\\
 \frac{dy}{dx} = \frac{2x-2y}{2x-2y+1} \\\\


klo yang turunan kedua itu
 \frac{d^2y}{dx^2}= \frac{d}{dx}(y')  \\\\
\frac{d}{dx}(y') =\frac{d}{dx}(\frac{2x-2y}{2x-2y+1} ) \\\\
 \frac{d^2y}{dx^2} =\frac{(2x-2y)'(2x-2y+1)-(2x-2y)(2x-2y+1)'}{(2x-2y+1)^2}

 \frac{d^2y}{dx^2} =\frac{(2-2.y')(2x-2y+1)-(2x-2y)(2-2.y')}{(2x-2y+1)^2}

 \frac{d^2y}{dx^2} =\frac{(4x-4y+2-4x.y'+4y.y'-2.y')-(4x-4x.y'-4y+4y.y')}{(2x-2y+1)^2}

 \frac{d^2y}{dx^2} =\frac{2-2.y'}{(2x-2y+1)^2}

 \frac{d^2y}{dx^2} =\frac{2-2(\frac{2x-2y}{2x-2y+1})}{(2x-2y+1)^2}

 \frac{d^2y}{dx^2} =\frac{2-2(\frac{2x-2y}{2x-2y+1})}{(2x-2y+1)^2}\times  \frac{(2x-2y+1)}{(2x-2y+1)} \\\\
 \frac{d^2y}{dx^2} =\frac{2(2x-2y+1)-2(2x-2y)}{(2x-2y+1)^3}

\frac{d^2y}{dx^2} =\frac{2(2x-2y+1)-2(2x-2y)}{(2x-2y+1)^3}\\\\
\frac{d^2y}{dx^2} =\frac{2}{(2x-2y+1)^3}
mungkin ada cara yg lebih singkat, tp sy taunya cuma kyk diatas