What is the implicit derivative of #1=xy^2-y^2+x^2 #?

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Answer 1 · 2017-02-01T19:41:18

#(dy)/(dx) = -(2x + y^2)/(2xy - 2y)#

Assuming #y# is a function of #x# and not the opposite:

Differentiate both sides with respect to #x#.

#(x^2 - y^2 + xy^2)' = (1)'#

#2x - 2y(dy)/(dx) + y^2 + 2xy(dy)/(dx) = 0#

#(dy)/(dx)(2xy - 2y) = -(2x + y^2)#

#(dy)/(dx) = -(2x + y^2)/(2xy - 2y)#

(If #f# is a function of #y#, then #(df)/(dx) = (df)/(dy) (dy)/(dx)#)