Is $e^{|x|}$ differentiable? My thoughts go as follows:
For $x > 0$, $e^{|x|} = e^x $
For $x < 0$, $e^{|x|} = e^{-x}$
Both $e^x$ and $e^{-x}$ are differentiable at every point in their domains, so $e^{|x|}$ will be differentiable for all $x \ne 0$
$e^{|x|}$ is certainly continuous everywhere, so I can't rule out differentiability with that criterion. 
I know the derivative of $e^x$ at $x=0$ is $1$, and the derivative of $e^{-x}$ at $x = 0$ is $-1$, so to me this indicates that the right hand limit and left hand limit of 
$\frac{e^{|x|} - 1}{x}$ approach different values as $x$ approaches $0$, so it cannot be differentiable at $0$.
This seems logically correct to me, but I'm not completely certain, and it feels a little weak. Any advice?
 A: You have a decent idea; however, the derivative of a function is not guaranteed to be continuous, so the argument doesn't work as is. Here is a way to fix it.
By Darboux's theorem, if $g$ is the derivative of a function, then $g$ has the intermediate value property.
Away from zero, the derivative of $f(x) = e^{|x|}$ is given by
$$ f'(x) = \operatorname{sign}(x) e^{|x|} $$
It's fairly easy to see that this function doesn't attain any values in the interval $[-1, 1]$ for nonzero $x$.
However, since $f$ does have values larger than $1$ and smaller than $-1$, the intermediate value property implies every value in $[-1, 1]$ has to be attained someplace.
But the only possible place is at $x=0$, and $f'(0)$ cannot be every number in $[-1,1]$ simultaneously!
Thus, $f$ is not differentiable.
A: Yes, you are right. Note that the function $y=e^{|x|}$ is symmetric with respect to the $y$axis so at $x=0$ we have a cusp because the tangent of $e^x$ at this point is not null. 
A: NB: The limit I'm referring to isn't $\lim_{x\to 0^{\pm}} f'(x)$ as some commenters thought. Rather, it's $\lim_{h\to 0^{\pm}}\frac{f(0+h)-f(0)}{h}$

Your logic at the end is correct. If the function were to be differentiable, then necessarily the left and right limits exist and agree, so we can check that. The left-hand limit is the derivative of $e^{-x}$ at $0$ and the right hand limit is the derivative of $e^x$ at $0$. At zero the former is $-1$ and the later is $1$ so the limit doesn't exist, and the function isn't differentiable.
A: Let $f(x)=e^{|x|}$ and $g(x)=e^x$.
Then $f(x)=g(x)$ for $\color{blue}{x \ge 0}$, so
$$
f'_+(0)
= \lim_{h \to 0^+} \frac{f(h)-\color{blue}{f(0)}}{h}
= \lim_{h \to 0^+} \frac{g(h)-\color{blue}{g(0)}}{h}
\overset{(*)}{=} \lim_{h \to 0} \frac{g(h)-g(0)}{h}
= g'(0)
.
$$
The equality marked with $(*)$, where $h \to 0^+$ was replaced with $h \to 0$, uses that we know already that $g$ is a differentiable function, so that the limit defining $g'(0)$ exists (and hence is equal to the corresponding left and right limits).
So yes, your conclusion about the right-hand derivative is correct: indeed we have $f'_+(0)=g'(0)$. But it depends on $f(x)=g(x)$ being true for $x \ge 0$ and not just for $x>0$; otherwise it wouldn't be possible to substitute $f(0)=g(0)$ in the computation above.
Similarly, your conclusion about the left-hand derivative is also correct, but again it's crucial that $e^{|x|}=e^{-x}$ for $x \le 0$ and not just for $x<0$.
(So if you would get this question on a test, you would have to be very careful with the motivation, to convince the grader that you really know why you are allowed to do this!)
See also this answer.
