How do you implicitly differentiate #-1=-y^2x-2xy-ye^y #?

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Answer 1 · 2016-10-23T13:34:29

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

So firstly differentiate all parts individually

#-1=y^2x-2xy-ye^y#
#-1 => 0#
#-y^2x =>-2yxdy/dx+y^2#
#-2xy=>-2dy/dx-2y#
#-ye^y=>-e^ydy/dx-ye^ydy/dx#

(this is done using mostly the product rule i.e.
#y=f(x)g(x)#
#y'=f'(x)g(x)+f(x)g'(x)#)

leaving us with,

#0=-2yx dy/dx-y^2-2y-2xdy/dx-e^ydy/dx-ye^ydy/dx#

#2yx dy/dx +2xdy/dx+e^ydy/dx+ye^ydy/dx=-y^2-2y#

#dy/dx(2yx+2x+e^y+ye^y)=-y^2-2y#

#dy/dx=(-y^2-2y)/(2yx+2x+e^y+ye^y)#