Answer
The equation of the tangent line $l$ to the given curve at $(0,\frac{1}{2})$ is $$(l): y=\frac{1}{4}x+\frac{1}{2}$$
Work Step by Step
$$y=f(x)=\frac{1+x}{1+e^x}$$
1) Find $f'(x)$ $$f'(x)=\frac{(1+x)'(1+e^x)-(1+x)(1+e^x)'}{(1+e^x)^2}$$
$$f'(x)=\frac{1\times(1+e^x)-e^x(1+x)}{(1+e^x)^2}$$
$$f'(x)=\frac{1+e^x-e^x-e^xx}{(1+e^x)^2}$$
$$f'(x)=\frac{1-e^xx}{(1+e^x)^2}$$
2) The slope of the tangent line $l$ to the given curve at point $A(a,b)$ is $f'(a)$.
Therefore, at $(0,\frac{1}{2})$, the slope of tangent line $l$ is $$f'(0)=\frac{1-e^0\times0}{(1+e^0)^2}$$
$$f'(0)=\frac{1-0}{(1+1)^2}=\frac{1}{4}$$
The equation of the tangent line $l$ to the given curve at $(0,\frac{1}{2})$ has the following form: $$(l): (y-\frac{1}{2})=f'(0)(x-0)$$
$$(l): y=\frac{1}{4}x+\frac{1}{2}$$
