# How do you find an equation of the tangent line to the curve #xe^y+ye^x = 1# at the point (0,1)?

Factoring, collecting like terms and tidying up.....

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To find the equation of the tangent line to the curve xe^y + ye^x = 1 at the point (0,1), we can use the concept of implicit differentiation.

First, we differentiate both sides of the equation with respect to x:

d/dx(xe^y) + d/dx(ye^x) = d/dx(1)

Using the product rule and chain rule, we get:

e^y + xe^y(dy/dx) + ye^x + e^x(dy/dx) = 0

Next, we substitute the coordinates of the given point (0,1) into the equation:

e^1 + 0e^1(dy/dx) + 1e^0 + e^0(dy/dx) = 0

Simplifying this equation, we have:

e + dy/dx + 1 + dy/dx = 0

Combining like terms, we get:

2(dy/dx) + e + 1 = 0

Finally, we solve for dy/dx:

dy/dx = -(e + 1)/2

Therefore, the equation of the tangent line to the curve at the point (0,1) is:

y - 1 = -(e + 1)/2 * (x - 0)

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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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