How do you implicitly differentiate $- 4 = {(2 x + 3 y)}^{2} - y^{2} x$ $-4=(2x+3y)^2-y^2x$ ?

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Answer 1 · 2016-05-01T02:04:38

$y'=(y^2-12y-8x)/(2(6x+9y-xy))$

We start from the given equation and differentiate both sides of the equation with respect to x

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

$d/dx(-4)=d/dx((2x+3y)^2-y^2x)$

$0=2(2x+3y)^(2-1)*d/dx(2x+3y)-d/dx(y^2x)$

$0=2(2x+3y)*(2+3y')-[y^2*1+x*2y*y'])$

Simplify at this point to obtain the $y'$

$0=(4x+6y)*(2+3y')-y^2-2xyy'$

$0=(8x+12y+12xy'+18yy')-y^2-2xyy'$

$0=8x+12y+12xy'+18yy'-y^2-2xyy'$

$(y^2-12y-8x)=(12x+18y-2xy)y'$

$y'=(y^2-12y-8x)/(12x+18y-2xy)$

$y'=(y^2-12y-8x)/(2(6x+9y-xy))$

God bless....I hope the explanation is useful.

How do you implicitly differentiate -4=(2x+3y)^2-y^2x −4=(2x+3y)2−y2x-4=(2x+3y)^2-y^2x ?