What is the slope of the tangent line of #sqrt(ye^(x-y))= C #, where C is an arbitrary constant, at #(-2,1)#?

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Answer 1 · 2016-01-10T12:14:55

#slope =-1/0=-oo#

the given #sqrt(y*e^(x-y))=C# at #(-2,1)#

squaring both sides of the equation it becomes

#y*e^(x-y)=c^2#

differentiation both sides of the equation

#y'e^(x-y)+y*(e^(x-y))(1-y')=0#

#y'e^(x-y)+ye^(x-y)-ye^(x-y)*y'=0#

#y'(e^(x-y)-ye^(x-y))=-ye^(x-y)#

#y'=(-y(e^(x-y)))/((e^(x-y))(1-y))#

#y'=-y/(1-y)#

Evaluating #y'# at #y=1#

#y' = -1/(1-1)=-1/0=-oo#