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

The slope is

Now differentiate implicitly.

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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).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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