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 $C=1/(e-1/e) = e/(e^2-1)$ .

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