How do you differentiate $4x^2 +3y^2 -2xy = 32$ ?

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Answer 1 · 2015-06-27T16:15:33

$y'=((y-4x))/((3y-x))$

$4x^2+3y^2-2xy=32$

$D(4x^2+3y^2-2xy)=D(32)$

$8x+6yy'-2(xy'+y)=0$

Now solve for y' $rArr$

$8x+6yy'-2xy'-2y=0$

$8x+y'(6y-2x)-2y=0$

$y'(6y-2x)-2y=-8x$

$y'(6y-2x)=2y-8x$

$y'=((2y-8x))/((6y-2x))$

$y'=(cancel(2)(y-4x))/(cancel(2)(3y-x))$

$y'=((y-4x))/((3y-x))$

How do you differentiate 4x^2 +3y^2 -2xy = 324x^2 +3y^2 -2xy = 32?