I was wondering if such a function exist. I'm comfortable with derivatives of polynomial functions, and some other basic functions, but I'm wondering if there could exist a very complicated function that doesn't have a derivative.

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    $\begingroup$ Google "Weierstrass function". $\endgroup$ – David Mitra Oct 3 '15 at 11:49
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    $\begingroup$ Or, more simply, the absolute value function $|x|$ at $x=0$. $\endgroup$ – joriki Oct 3 '15 at 11:53
  • $\begingroup$ Related: Are Continuous Functions Always Differentiable? (And other posts linked there might be also of interest.) $\endgroup$ – Martin Sleziak Oct 3 '15 at 13:01
  • $\begingroup$ @mick, if my answer was useful, consider accepting it by clicking the gray tick on the left of it. $\endgroup$ – dbanet Oct 19 '15 at 18:08

The commonly known example is the Weierstrass Function $f$, defined as $$f(x)=\sum\limits_{k=1}^\infty\frac{\sin\left(\pi{k}^2x\right)}{\pi{k}^2}.$$

The $y=f(x)$ graph looks like this (and intuitively shows why it is differentiable nowhere):

Graph of the Weierstrass Function

Another example would be the Dirichlet Function $D$, defined as $$D(x)=\begin{cases}1,\;\;x\in\mathbb{Q},\\0,\;\;x\in\mathbb{I}.\end{cases}$$

Its graph $y=D(x)$ would look like a pair of lines $\displaystyle{y=\frac{1}{2}\pm\frac{1}{2}}$ (which of cousre is not a graph of any function), so is uninteresting to show. The interesting part is that $D$ is actually discontinuous everywhere, and therefore differentiable nowhere.

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    $\begingroup$ It's worth mentioning that the Weierstrass function is continuous everywhere. $\endgroup$ – joriki Oct 3 '15 at 12:10
  • $\begingroup$ does the Weierstrass function pass the vertical line test? $\endgroup$ – set5 Oct 3 '15 at 13:05
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    $\begingroup$ @mick Hint: It's a function. $\endgroup$ – Workaholic Oct 3 '15 at 13:18
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    $\begingroup$ @mick, any function passes the vertical line test. If you meant to ask about the Dirichlet Function $D$ instead, the source of confusion is clear: its graph only looks like a pair of parallel lines, but actually for every $x_0$ the line $x=x_0$ crosses only one of the “lines” of the graph (and “goes through a hole in the second one”, if that adds intuition). $\endgroup$ – dbanet Oct 3 '15 at 13:18

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