Given:
The equation of curve is:
[tex]y=\dfrac{2e^x}{1+e^x}[/tex]
To find:
The equation of the tangent line to the given curve at (0,1).
Solution:
We know that the slope of the tangent line at (a,b) is
[tex]m=\dfrac{dy}{dx}_{(a,b)}[/tex]
We have,
[tex]y=\dfrac{2e^x}{1+e^x}[/tex]
Differentiate with respect to x.
[tex]\dfrac{dy}{dx}=\dfrac{(1+e^x)(2e^x)'-2e^x(1+e^x)'}{(1+e^x)^2}[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{(1+e^x)(2e^x)-2e^x(e^x)}{(1+e^x)^2}[/tex]
Slope of the tangent is
[tex]\dfrac{dy}{dx}_{(0,1)}=\dfrac{(1+e^0)(2e^0)-2e^0(e^0)}{(1+e^0)^2}[/tex]
[tex]\dfrac{dy}{dx}_{(0,1)}=\dfrac{(1+1)(2(1))-2(1)(1)}{(1+1)^2}[/tex]
[tex]\dfrac{dy}{dx}_{(0,1)}=\dfrac{(2)(2)-2}{(2)^2}[/tex]
[tex]\dfrac{dy}{dx}_{(0,1)}=\dfrac{4-2}{4}[/tex]
on further simplification, we get
[tex]\dfrac{dy}{dx}_{(0,1)}=\dfrac{2}{4}[/tex]
[tex]\dfrac{dy}{dx}_{(0,1)}=\dfrac{1}{2}[/tex]
The slope of the tangent line is [tex]m=\dfrac{1}{2}[/tex] and it passes through the point (0,1). So, the equation of the tangent line is
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-1=\dfrac{1}{2}(x-0)[/tex]
[tex]y-1=\dfrac{1}{2}x[/tex]
[tex]y=\dfrac{1}{2}x+1[/tex]
Therefore, the equation of the tangent line is [tex]y=\dfrac{1}{2}x+1[/tex].