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

1 Answer
Feb 18, 2017

graph{(y - (2e^x)/(1+e^x)) (y- (1/2x + 1)) = 0 [-10, 10, -5.21, 5.21]}

The equation of the tangent line at (0,1) is y=1/2x +1

Explanation:

Consider x = 0 , y = 1

Using Quotient Rule of Differentiation:

Let u = 2e^x and v = 1 + e^x

dy/dx = (vu' - uv')/v^2 , and given that d e^x/dx = e^x

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

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

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

When x = 0

dy/dx = [(2e^(0))/(1+e^0)] - {([(2e^0)] [e^(0)])/((1+e^0)^2)}

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

dy/dx = 2/2 -2/2^2

dy/dx = 1 - 1/2

dy/dx = 1/2

So the slope at x = 0 is 1/2 (f'(x) = m = 1/2).

Consider:

y - y_0 = m(x-x_0)

y_0 = 1, x_0 = 0 ( this is at the point (0,1) )

y - f(x_0) = f'(x_0)(x-x_0)

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

y = 1/2x +1

So the equation of the tangent line at (0,1) is y=1/2x +1