How do you differentiate #2xy+y^2=x+y#?

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Answer 1 · 2015-08-14T10:09:27

2xy + #y^2# = x + y

By transpose we get

2xy + #y^2# - x - y = 0

Use product rule to differentiate 2xy

2x . #dy/dx# + 2y . 1 + 2y #dy/dx# - 1 - #dy/dx# = 0

2x . #dy/dx# + 2y + 2y #dy/dx# - 1 - #dy/dx# = 0

Rearrange
2x . #dy/dx# + 2y #dy/dx# - #dy/dx# + 2y - 1 = 0

2x . #dy/dx# + 2y #dy/dx# - #dy/dx# = - 2y + 1

#dy/dx# (2x + 2y - 1) = - 2y + 1

#dy/dx# = #( - 2y + 1)/ (2x + 2y - 1) #