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

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Answer 1 · 2018-07-23T14:44:32

The answer is #=-(y^2+2y-ye^x)/(2yx+2x-e^x)#

Let

#f(x,y)=y^2x+2xy-ye^x-1#

Calculate the partial derivatives

#(delf)/(delx)=y^2+2y-ye^x#

#(delf)/(dely)=2yx+2x-e^x#

Therefore,

#dy/dx=-((delf)/(delx))/((delf)/(dely))#

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