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

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Answer 1 · 2015-08-14T09:55:45

#x^2#+2xy−#y^2#+x=2

To differentiate 2xy use product rule

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

Rewrite it

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

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

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

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