Derivative of a function with modulus terms. Please any one help me in calculation of $f(x)$, where $f(x)=-i\hbar\frac{\partial}{\partial x}\left(e^{-\frac{\left|x\right|}{a}}\right)$. Here $\hbar$ and $a$ is constant. Also scetch the $f(x)$.
 A: $f(x)=-i\hbar\frac{\partial}{\partial x}\left(e^{-\frac{\left|x\right|}{a}}\right)$
It is sufficient to apply chain rule.
$\newcommand{\sign}{\text{sign }}$
$f(x)=-i\hbar e^{-\frac{\left|x\right|}{a}}\cdot\left(-\frac{1}{a}\sign x\right)=i\frac{\hbar}{a}(\sign x)e^{-\frac{\left|x\right|}{a}}\\\forall x\in\mathbb{R}\land x\ne0$
where $\;\sign x= \left \{ \begin{array}{rl}
1\;\;\;\;\;\text{ if }x>0\\
-1\;\;\;\;\;\text{ if }x<0
\end{array}
\right.$
$f(x)$ is not defined at $x=0$ and
$\lim_\limits{x\to0^-}f(x)=-i\frac{\hbar}{a}$ ,
$\lim_\limits{x\to0^+}f(x)=i\frac{\hbar}{a}$ .
The graph of the function $f(x)$ for $a=1.86$ is the following:

A: Let, $\psi(x)=e^{-\frac{\left|x\right|}{a}}$. So according to the question,
$$f(x)=\widehat{P}_x\psi(x)=-i\hbar \frac{\partial}{\partial x}\left(e^{-\frac{\left|x\right|}{a}}\right)$$
Now we can consider that $\left|x\right|=\sqrt{x^2}$. By considering this identity $f(x)$ can be rewritten as,
$$f(x)=-i\hbar \frac{\partial}{\partial x}\left(e^{-\frac{\sqrt{x^2}}{a}}\right)$$
Now it is sufficient to apply chain rule for the calculation of $f(x)$-
$$f(x)=-i\hbar \cdot \left(-\frac{1}{a}\right)\cdot \left(e^{-\frac{\sqrt{x^2}}{a}}\right)\cdot \left(\frac{1}{2\sqrt{x^2}}\right)\cdot \left(2x\right)$$
$$\Rightarrow f(x)=\frac{i\hbar}{a}\cdot\frac{x}{\sqrt{x^2}}\cdot \left(e^{-\frac{\sqrt{x^2}}{a}}\right)$$
$$\Rightarrow f(x)=\frac{i\hbar}{a}\cdot\frac{x}{\left|x\right|}\cdot \left(e^{-\frac{\left|x\right|}{a}}\right)$$
So, $f(x)$ exists for $\forall x \in \mathbb{R} \wedge x \neq 0$
So, $f(x)$ is not defined only at $x=0$ and $\displaystyle{\lim_{x \to 0^+}} f(x)=\frac{i\hbar}{a}$, $\displaystyle{\lim_{x \to 0^-}} f(x)=-\frac{i\hbar}{a}$.
Now we will consider a variable transformation that is $x'=ix$, where $x'$ is a purely imaginary variable. So,
$$f(x')=\frac{\hbar}{a}\cdot\frac{x'}{\left|x'\right|}\cdot \left(e^{-\frac{\left|x'\right|}{a}}\right)$$
Now, a graphical representation of $f(x')$ vs. $x'$ is following-

