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

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.