How do you find the equation of the tangent to the curve defined by #y= (2e^x) / (1+e^x)# at the point (0,1)?

To find the equation of the tangent to the curve at the point (0,1), we need to find the slope of the tangent line. The slope of the tangent line is equal to the derivative of the function at that point.

First, let's find the derivative of the function y = (2e^x) / (1+e^x). Using the quotient rule, we have:

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

Simplifying this expression, we get:

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

dy/dx = (2e^x) / (1+e^x)^2

Now, let's find the slope of the tangent line at the point (0,1) by substituting x = 0 into the derivative:

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

dy/dx = 2 / (1+1)^2

dy/dx = 2 / 4

dy/dx = 1/2

Therefore, the slope of the tangent line at the point (0,1) is 1/2.

Using the point-slope form of a line, we can write the equation of the tangent line as:

y - 1 = (1/2)(x - 0)

Simplifying this equation, we get:

y - 1 = 1/2x

y = 1/2x + 1

Hence, the equation of the tangent to the curve at the point (0,1) is y = 1/2x + 1.