How do you find #dy/dx# by implicit differentiation given #2xy^-2+x^-2=y#?
1 Answer
Dec 23, 2017
Explanation:
#"differentiate "2xy^-2" using the "color(blue)"product rule"#
#2(-2xy^-3.dy/dx+y^-2)-2x^-3=dy/dx#
#rArr-4xy^-3dy/dx+2y^-2-2x^-3=dy/dx#
#rArrdy/dx(1+4xy^-3)=2x^-3-2y^-2#
#rArrdy/dx=(2x^-3-2y^-2)/(1+4xy^-3)#
