Derivative?

Clearly the point is (1,1)=>eqn of tangent is y-1=m(x-1) which gives y= mx+(1-m).Hence m+n=1.Where is the need to evaluate dy/dx at x=1 :)

$y = x^{x^{x^{x^{.^{.^{.}}}}}} \quad \Rightarrow y = x^y \quad \Rightarrow \ln{y} = y \ln{x}$

$\begin{aligned} \Rightarrow \frac{d}{dx} (\ln{y}) & = \frac{d}{dx} \left(y \ln{x} \right) \\ \frac{1}{\color{#3D99F6}{y}} \frac{dy}{dx} & = \ln{x} \frac{dy}{dx} + \frac{\color{#3D99F6}{y}}{x} \quad \quad \color{#3D99F6}{[x = 1 \quad \Rightarrow y = 1^y \quad \Rightarrow y =1 ]} \\ \frac{1}{\color{#3D99F6}{1}} \frac{dy}{dx} & = \ln{1} \frac{dy}{dx} + \frac{\color{#3D99F6}{1}}{1} \\ \Rightarrow \frac{dy}{dx} & = \color{#D61F06}{1} = \color{#D61F06}{m} \end{aligned}$

$\begin{aligned} \Rightarrow \frac{y-1}{x-1} & = \color{#D61F06}{m} = \color{#D61F06}{1} \\ \Rightarrow y -1 & = x - 1 \\ \Rightarrow y & = \color{#D61F06}{1}x + \color{#D61F06}{0} \\ \Rightarrow \color{#D61F06}{m} + \color{#D61F06}{n} & = \color{#D61F06}{1} + \color{#D61F06}{0} = \boxed{1} \end{aligned}$

We start with $y=x^{x^{x^{x^{.^{.^{.}}}}}}$

Taking the natural log of both sides we have: $\ln y=\ln x^{x^{x^{x^{.^{.^{.}}}}}} = x^{x^{x^{x^{.^{.^{.}}}}}} \ln x = y \ln x \implies \frac{\ln y}{y}=\ln{x}$

By implicit differentiation: $\frac{1+\ln y}{y^2} \frac{dy}{dx} =\frac{1}{x}$

At $x=1$ , $y=1^{1^{1^{1^{.^{.^{.}}}}}}=1$ so $\frac{1+\ln 1}{1^2} \frac{dy}{dx} =\frac{1}{1} \implies \frac{dy}{dx}=1$

We now have $m=1$ and point $(1,1)$ on the function at $x=1$ so $y=1(x-1)+1=x$ which gives $m=1, n=0,$ and the answer is $1+0=\boxed{1}$

$m + n = m\cdot 1 + n = mx + n = y = x^{x^{x^{\dots}}} = 1^{1^{1^{\dots}}} = \boxed{1}.$