Why use the derivative and not the symmetric derivative? The symmetric derivative is always equal to the regular derivative when it exists, and still isn't defined for jump discontinuities. From what I can tell the only differences are that a symmetric derivative will give the 'expected slope' for removable discontinuities, and the average slope at cusps. These seem like extremely reasonable quantities to work with (especially the former), so I'm wondering why the 'typical' derivative isn't taken to be this one. What advantage is there to taking $\lim\limits_{h\to0}\frac{f(x+h)-f(x)} h$ as the main quantity of interest instead? Why would we want to use the one that's defined less often?

Following my comment on the Mean Value Theorem.  Since MVT fails, anything we prove from MVT is likely to fail as well.  For example:
Find the minimum of the (symetrically) differentiable function
$f(x) = x+2|x|$ on the interval $[-1,1]$.
Usual solution: find where the derivative is zero.  Answer: nowhere!
Since $f'(x) = -1$ on $[-1,0)$, $f'(0)=1$,
and $f'(x)=3$ on $(0,1]$.
