What is the implicit derivative of #1= xy-e^(xy) #?

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Answer 1 · 2016-06-21T17:47:16

#y' = - y/x, \qquad x,y \ne 0#

# 0 = d/dx (xy) - d/dx e^{xy}#

first term, using product rule:

# d/dx (xy) = y + x y'#

second term using chain rule:

#d/dx e^{xy} = d/dx(xy) * e^{xy}#

with product rule this is #d/dx(xy) * e^{xy} = (y + x y') * e^{xy}#

combining terms

#d/dx (xy) - d/dx e^{xy} = (y + x y') - (y + x y') * e^{xy} #

#0 = (y + x y') ( 1 - e^{xy})#

ignoring trivial solution #( 1 - e^{xy}) = 0 \implies x = 0, or y = 0# leaves

# y + x y' = 0 \implies y' = - y/x, \qquad x,y \ne 0#