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

Question

William Abdalla · Accepted Answer

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)#.

Emily Ammerman · Answer

undefined To find the slope of the tangent line at a point on the curve $ \frac{1}{e^y - e^x} = C $, we need to take the derivative implicitly with respect to $x$ and $y$, then evaluate it at the point (-1,1).

The implicit differentiation yields:

$$ \frac{-e^y}{(e^y - e^x)^2} \frac{dy}{dx} + \frac{e^x}{(e^y - e^x)^2} = 0 $$

Solving for $ \frac{dy}{dx} $:

$$ \frac{dy}{dx} = \frac{e^x}{e^y - e^x} $$

Substitute (-1,1):

$$ \frac{dy}{dx} = \frac{1}{e^1 - e^{-1}} = \frac{1}{e - \frac{1}{e}} = \frac{e^2}{e^2 - 1} $$

This is the slope of the tangent line at (-1,1).