I already know that there are functions that have no anti-derivative (although I haven't seen a proof), but is there a function with no derivative? and I don't mean something like dirichlet function or floor/ceiling functions. In other words, is there some function that is defined in terms of algebraic operations that have no derivative?
(just for reference I've taken calc I,II,III,linear algebra and am currently in diff eq's)
edit: by algebraeic operations what i really meant is "closed-form expression"
I understand that you mean a derivative with no closed form formula.
No, there aren't such functions, as there is an algorithm to compute the symbolic derivative.
Any closed form expression can be expressed as a syntax tree (f.i. $\cos(x^2+3)$, is described by a $\cos$ node with a son; the son is a sum ($+$) node with two sons; the left son is a square node with a son, which is the independent variable $x$; the right son is the constant $3$).
And every kind of node corresponds to a derivation rule, where every closed formula derives as another closed formula. $$(u+v)'=u'+v'$$ $$(uv)'=u'v+uv'$$ $$(\cos(u))'=-u'\sin(u)$$ $$(e^u)'=u'e^u$$ $$\cdots$$
By induction, the result is a closed formula.
If you allow special functions, such as $\Gamma(x)$, then you have to also allow their derivatives, which may not be expressible in terms of the original function.
Or said differently, if you allow the $\Gamma$ function alone, then, yes, there are functions without a derivative.