How do you differentiate #2xy=y^2-x^2/y#?
1 Answer
Nov 16, 2016
# dy/dx = (2(y^2 - x))/(y(3y-4x)) #
Explanation:
As we will differentiate implicitly we may as well multiply through by
# 2xy = y^2-x^2/y #
# :. 2xy^2 = y^3-x^2 #
We now differentiate wrt
# (2x)(d/dxy^2) + (d/dx2x)(y^2) = d/dxy^3 - 2x #
# :. (2x)(2ydy/dx) + (2)(y^2) = 3y^2dy/dx - 2x #
# :. 4xydy/dx + 2y^2 = 3y^2dy/dx - 2x #
# :. 3y^2dy/dx-4xydy/dx = 2y^2 - 2x #
# :. (3y^2-4xy)dy/dx = 2(y^2 - x) #
# :. dy/dx = (2(y^2 - x))/(y(3y-4x)) #
