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

1 Answer

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

Explanation:

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.