How do you find #(dy)/(dx)# given #2xy+y^2=x+y#?

2 Answers
Jun 11, 2018

#\frac{1-2y}{2x+2y-1}#

Explanation:

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

Use product rule and finish it off to get:

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

to make it clearer, let #\frac{d}{dx}=y'#

so you get:
#2(xy'+y)+2yy'=1+y'#

Rearrange to make #y'# the factor to get:

#y'=\frac{1-2y}{2x+2y-1}#

bring back #y'=\frac{d}{dx}#

so you get:
#\frac{d}{dx}: \frac{1-2y}{2x+2y-1}#

Jun 11, 2018

#dy/dx=(1-2y)/(2x+2y-1)#

Explanation:

#color(blue)"differentiate implicitly with respect to x"#

#"noting that "d/dx(y)=dy/dx" and "d/dx(y^2)=2ydy/dx#

#"differentiate "xy" using the "color(blue)"product rule"#

#2xdy/dx+2y+2ydy/dx=1+dy/dx#

#dy/dx(2x+2y-1)=1-2y#

#dy/dx=(1-2y)/(2x+2y-1)#