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

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Answer 1 · 2016-01-06T16:17:25

The slope is #1/e^2 = e^(-2)# and #C# is not arbitrary, it is #e/(e^2 -1)#.

Given #1/(e^y-e^x) = C # for constant #C#.

We can rewrite this as #e^y-e^x=1/C#.

Now differentiate implicitly.

#e^y dy/dx -e^x=0#. So,

#dy/dx = e^x/e^y#

Given that the point #(-1,1)# lies on the graph of this equation, we have

#dy/dx ]_"(-1,1)" = e^(-1)/e^1 = 1/e^2#

And #(-1,1)# being on the graph also determines #C# for this graph.
Since #1/(e^y-e^x) = C #, we find that #C=1/(e-1/e) = e/(e^2-1)#.