Is this function twice differentiable at $0$? I have a function $f(x)$:
$$f(x)=\frac{\exp(-|x|)}{1-0.5|\tanh(2x)|}$$
If I try differentiating it in Mathematica (taking $|x|=(2\theta(x)-1)x$ where $\theta(x)$ is Heaviside step function), I get answer in terms of Heaviside functions. But looking at the derivative I can't really see any jump there, which is why I start thinking that it has a second derivative too. But differentiating once more, I get answer with Dirac functions, which don't disappear when evaluating the result at $x=0$.
If I instead represent $|x|=\sqrt{x^2}$, then for the first derivative I get an expression which has $0$ as its limit as $x\to0$, and for second derivative the limit appears to be 1. So from this I could make the conclusion that $f(x)$ is twice differentiable at $x=0$.
So the question: is this function differentiable at $x=0$? Is it twice differentiable?
 A: Since we are dealing with real $x$ (we are, aren't we?), we can write
$$\lvert \tanh (2x)\rvert = \tanh (2\lvert x\rvert).$$
We have $\frac12\tanh (2\lvert x\rvert) = \lvert x\rvert + O(\lvert x\rvert^3)$, so we can calculate
$$\begin{align}
\frac{e^{-\lvert x\rvert}}{1 - \frac12\tanh(2\lvert x\rvert)} &= \left(1 - \lvert x\rvert + \frac{\lvert x\rvert^2}{2} + O(\lvert x\rvert^3)\right)\left(1 + \lvert x\rvert + \lvert x\rvert^2 + O(\lvert x\rvert^3)\right)\\
&= 1 + \frac{\lvert x\rvert^2}{2} + O(\lvert x\rvert^3).
\end{align}$$
Since the two halves of the function are analytic, it follows that the function is twice continuously differentiable.
Generally, if we have a function of the form
$$f(x) = \begin{cases}g(x) &, x \geqslant 0\\ h(x) &, x \leqslant 0 \end{cases}$$
where $g$ and $h$ are continuously differentiable functions with $g(0) = h(0)$ - so that the definition of $f$ has no conflict - then $f$ is differentiable in $0$ if and only if $g'(0) = h'(0)$, and then $f$ is continuously differentiable with $f'(0) = g'(0) = h'(0)$.
If $g$ and $h$ are analytic, i.e. we have power series representations
$$g(x) = \sum_{n=0}^\infty a_n x^n,\qquad h(x) = \sum_{n=0}^\infty b_n x^n,$$
then $f$ is $k$ times (continuously) differentiable if and only if $a_n = b_n$ for $0 \leqslant n \leqslant k$. The computation above shows that the two functions
$$g(x) = \frac{e^{-x}}{1-\frac12\tanh(2x)}\quad\text{and}\quad h(x) = \frac{e^x}{1+\frac12\tanh(2x)}$$
both have power series expansions starting with $1 + \frac12 x^2$, so $f$ is twice continuously differentiable.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
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$\large\tt Hint:$
\begin{align}
\verts{\tanh\pars{2x}} &= \tanh\pars{2\verts{x}}
\\[3mm]
\verts{\tanh\pars{2x}}\,'
&=
\sech^{2}\pars{2\verts{x}}
\bracks{2\sgn\pars{x}}
=
2\sgn\pars{x}\sech^{2}\pars{2x}
\\[5mm]
\verts{\tanh\pars{2x}}\,''
&=
2\delta\pars{x}\sech^{2}\pars{2x} + 2\sgn\pars{x}
\bracks{-4\tanh\pars{2x}\sech^{2}\pars{2x}}
\\[3mm]&=
2\delta\pars{x} - 8\tanh\pars{2\verts{x}}\sech^{2}\pars{2x}
\\[5mm]
\pars{\expo{-\verts{x}}}' &= -\sgn\pars{x}\expo{-\verts{x}}
\\[3mm]
\pars{\expo{-\verts{x}}}'' &= -2\delta\pars{x}\expo{-\verts{x}}
-
\sgn\pars{x}\expo{-\verts{x}}\bracks{-\sgn\pars{x}}
=
-2\delta\pars{x} + \expo{-\verts{x}}
\end{align}
