What is the slope of the tangent line of #x^2-ye^(x-y)= C #, where C is an arbitrary constant, at #(0,0)#?

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Answer 1 · 2018-06-27T23:44:24

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#x^2-ye^x/e^y=C# #0^2-0e^0=C# #C=0#

#e^y(x^2)=ye^x#

#color(blue)(d/dx)(e^y(x^2))=color(blue)(d/dx)(ye^x)#

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

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

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

#dy/dx(0,0)=(0e^0-0e^0)/(e^0-0e^0=0/1=0#