find an equation of the tangent line to the curve at the given point y=1+x/1+e^x (0,1/2)

Question

Please help me

Answer 1

The equation of a line is y = mx + b where m is the slope and b is the y-intercept.  So you need to find the slope (m) and y-intercept (b) of the tangent line.  The slope is easy - it's just the derivative of f(x) at the given point, (0,1/2); that is, m = df(x)/dx at x = 0:

 

     f(x) = (1+x)/(1+e^x)

 

Apply the quotient  and chain rules to find df(x)/dx:

 

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

 

     df(0)/dx = 1/4 = m, the slope of the tangent line to f(x) at x = 0. 

 

So the equation of the tangent line so far is:

 

     y = (1/4)x + b

 

To find b, plug in the given point (0,1/2) and solve for b:

 

     y = (1/4)x + b

     1/2 = 0 + b

 

     1/2 = b

 

So the equation of the tangent line is:

 

     y = (1/4)x + 1/2