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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9Google "Weierstrass function". – David Mitra Oct 03 '15 at 11:49
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5Or, more simply, the absolute value function $|x|$ at $x=0$. – joriki Oct 03 '15 at 11:53
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Related: Are Continuous Functions Always Differentiable? (And other posts linked there might be also of interest.) – Martin Sleziak Oct 03 '15 at 13:01
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@mick, if my answer was useful, consider accepting it by clicking the gray tick on the left of it. – dbanet Oct 19 '15 at 18:08
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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):
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.
dbanet
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16It's worth mentioning that the Weierstrass function is continuous everywhere. – joriki Oct 03 '15 at 12:10
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2@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). – dbanet Oct 03 '15 at 13:18