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

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Answer 1 · 2017-02-18T05:48:05

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#

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#