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

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### 1 Answer

To find the slope of the line tangent, we will need to differentiate the function.

`y=(2e^x)/(1+e^x) =>`

`y'=(2e^x(1+e^x)-2e^x(e^x))/(1+e^x)^2 =>`

`y'=(2e^x+2e^(2x)-2e^(2x))/(1+e^x)^2 =>`

`y'=(2e^x)/(1+e^x)^2`

Now we will evaluate that at the point (0,1), which means we will be plugging in 0 for x.

`y'(0)=(2e^0)/(1+e^0)^2=2/(1+1)^2=2/4=1/2`

Now we will find the equation of the line by using slope m=1/2 and the point (0,1)

`y-1=(1/2)(x-0) => y=(1/2)x+1`