Help with Derivatives of Exponential Functions?

Question

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

Answer 1

y = (2e^x) / (1 + e^x)
Quotient rule:
y' = [ (2e^x)(1 + e^x) - (2e^x)(e^x) ] / (1 + e^x)^2

Let x = 0 for the slope:
y' = [(2)(1 + 1) - (2)(1)] / (1 + 1)^2
y' = (4 - 2) / (4)
y' = 1/2

Eqn:

(y -y1) = m(x -x1)
y1 = 1, x1 = 0, m = 1/2

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

Answer 2

The derivative (with respect to x) of ke^(f(x)) is simply
kf'(x)e^(f(x)). And the derivative of the quotient of two functions f(x)/g(x) is simply (f'(x)g(g) - f(x)g'(x))/(g(x))². So the derivative you're looking for is
((2e^x)*(1+e^x) - (2e^x)*(e^x))/(1+e^x)² evaluated at x=0.
((2)*(1+1) - (2)*(1))/(2)² = 1/2


Doug

EDIT: Oops... I forgot the equation for the tangent line. But Mark H and AthiestGuy have it correct ☺

Answer 3

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

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

at the point (0,1):
y' = (2e°) / (1+e°)² = 2/2² = 2/4 = 1/2
=>
m = 1/2

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

y = x/2 + 1

₢

Answer 4

=(the origenal fuction)*(the derivatives of the upper part of the functions)

Answer 5

y = .5x + 1

or

2y - x = 2