The graph of y=e^tanx-2 crosses the x axis at one point in the interval [0,1]. what is the slope of the graph at this point? i get 3.08267 but it's not the right answer

Y = e^tanx   -   2

To find at which point it crosses x axis we state that y= 0

e^tanx   - 2 = 0 
e^tanx = 2
tanx = ln 2
tanx = 0.69314
x = 0.6061

to find slope at that point first we need to find first derivative of funtion y.

y' = (e^tanx)*1/cos^2(x)

now we express x = 0.6061 in y' and we get:
y' = k = 2,9599

What are the options