How do you differentiate #e^(xy) + x^2 +y^2 = 12#?

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Answer 1 · 2016-06-21T22:51:40

# y' = -(y \ e^(xy) + 2x)/(x\ e^(xy) + 2y )#

#d/(dx) { e^(xy) + x^2 +y^2 = 12}#

#\implies \color{red}{ (xy)'} e^(xy) + 2x +2y \ y' = 0#

red bit by chain rule

#\implies \color{blue}{(y + x y')} e^(xy) + 2x +2y \ y' = 0#

#\implies y*e^(xy) + x y' * e^(xy) + 2x +2y \ y' = 0#

#\implies y*e^(xy) + y' (xe^(xy) + 2y ) + 2x = 0#

#\implies y' = -(y \ e^(xy) + 2x)/(x\ e^(xy) + 2y )#