Find the slope of the tangent line to the curve (x + (e^-x))^2 at (0,1)

Answer 1

the slope of the tangent line is the derivative of y=(x+e^-x)^2 at x=0

y' = 2(x + e^-x)(1 - e^-x)
y'(0) = 2(0+1)(1-1) = 0