How do you find #dy/dx# by implicit differentiation of #x^3-3x^2y+2xy^2=12#?
1 Answer
Jan 10, 2017
Explanation:
To differentiate
#-3x^2y" and " 2xy^2# we use the#color(blue)"product rule"#
#rArr3x^2-(3x^2.dy/dx+6xy)+(2x.2ydy/dx+2y^2)=0#
#rArr3x^2-3x^2dy/dx-6xy+4xydy/dx+2y^2=0#
#rArrdy/dx(4xy-3x^2)=6xy-3x^2-2y^2#