How do you find #(dy)/(dx)# given #2xy+y^2=x+y#?
2 Answers
Jun 11, 2018
Explanation:
Use product rule and finish it off to get:
to make it clearer, let
so you get:
Rearrange to make
bring back
so you get:
Jun 11, 2018
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)#