y = 6/(1+e^-x) = 6e^x/(1+e^x)
y' = 6e^x/(1+e^x)^2
so, at (0,3) the slope is 3/2
Now you have a point and a slope, so the line is
y-3 = 3/2 (x-0)
see the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D+6%2F%281%2Be^-x%29%2C+y%3D3%2F2+x%2B3
Find an equation of the tangent line to the curve
y = 6/(1+e^−x)at the point (0,3).
1 answer