The graph of \(y=e^{\tan x}-2\) crosses the \(x\)-axis at one point in the interval \([0,1]\). What is the slope of the graph at this point?

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Find the \(x\) value for when the graph crosses the \(x\)-axis:

$$ 0 = e^{\tan x}-2 $$ $$ 2=e^{\tan x} $$ $$ \ln 2=\tan x $$ $$ x=\arctan (\ln 2) $$

Obtain the derivative:

$$ y=e^{\tan x}-2 $$ $$ y'=\sec^2 x \cdot e^{\tan x} $$

Finding slope with calculator:

$$ y'\Big|_{x=\arctan (\ln 2)} = \sec^2(\arctan(\ln 2))\cdot e^{\tan (\arctan (\ln 2)) } $$ $$ = 2\sec^2(\arctan(\ln 2)) $$ $$ \approx \boxed{2.961} $$