How do you use implicit differentiation to find dy/dx given #2x-y+y^2=0#?
1 Answer
Nov 13, 2016
#dy/dx=(-2)/(2y-1)#
Explanation:
Given-
#2x-y+y^2=0#
#2-dy/dx+2y*dy/dx=0#
#-dy/dx+2y*dy/dx=-2#
#dy/dx(-1+2y)=-2#
#dy/dx(2y-1)=-2#
#dy/dx=(-2)/(2y-1)#
