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

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Answer 1 · 2017-02-03T22:24:59

The slope at # (2,1)# is 1

#(y+e^x)/(y-e^y) =C# is a family of equations, and we are specifically interested in those family members that pass through # (2,1)#.

One approach is to proceed with the calculus, as this is posted as a claculus problem. We can do some simplifying algebra first:

#y+e^x =C(y-e^y) implies y(1-C) =-e^x-Ce^y#

If we now differentiate wrt x in order to obtain the slope:

#y'(1-C) =-e^x-y'Ce^y#

#implies y'(1 + C(e^y-1)) =-e^x#

#implies y' =-e^x/(1 + C(e^y-1))#, the slope

At # (-2,1)# that means that:

#y' =-1/(e^2(1 + C(e-1)))#

But we also know that #(1+e^(-2))/(1-e) =C# and so we can sub that into our slope equation to obtain:

#y' =-1/(e^2(1 + (1+e^(-2))/(1-e) (e-1)))#

# =-1/(e^2(1 - (1+e^(-2))))#

# =-1/(-1) color(blue)(= 1)#

This is how it occurred me to do it off the bat. I reckon there must be better/cleverer ways but I leave that to you!