# 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)#?

The slope at

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

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

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!

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To find the slope of the tangent line of the given curve at the point (-2,1), we first need to find the derivative of the curve with respect to x using implicit differentiation. Then, we evaluate the derivative at the point (-2,1) to find the slope of the tangent line.

Differentiating both sides of the equation (y + e^x) / (y - e^y) = C with respect to x, we get:

(dy/dx)(y - e^y) - (y + e^x)(dy/dx)(1 - e^y(dy/dx)) = 0

Solving for dy/dx, we get:

dy/dx = [(y - e^y)(e^y - 1) - (y + e^x)(1 - e^y(dy/dx))] / (y + e^x)

Now, we substitute the point (-2,1) into this expression to find the slope of the tangent line at that point. We need to find the values of y and dy/dx at x = -2.

Given the point (-2,1), we substitute x = -2 into the original equation to find y:

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

Substitute y = 1 into the equation:

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

Solve for C:

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

Now, substitute x = -2 and y = 1 into dy/dx:

dy/dx = [(1 - e)(e - 1) - (1 + e^(-2))(1 - e(dy/dx))] / (1 + e^(-2))

Evaluate the expression to find dy/dx at x = -2, y = 1, and then you'll have the slope of the tangent line at that point.

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To find the slope of the tangent line of the curve given by ( \frac{y+e^x}{y-e^y} = C ) at the point (-2,1), differentiate implicitly with respect to ( x ), then evaluate the derivative at the given point. After differentiating implicitly and solving for ( \frac{dy}{dx} ), substitute the coordinates of the point (-2,1) to find the slope of the tangent line at that point.

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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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