Like the title suggests. Is it possible to have an implicit function that is continuous but not differentiable? Something which resembles a fractal, or is perhaps constant (not asymptotic) after a certain x but without a smooth approach, describable implicitly in x and y. On a related note can one describe a self-similar function, like a fractal, implicitly? For example a sinusoidal with noise is often self similar and always continuous but not differentiable anywhere. I am not referring to solutions given by the Implicit Function theorem which maps relations to functions.


Consider $f:\mathbb{R}^{2}\rightarrow \mathbb{R}$ defined by $f(x,y)=y^{3}-x$

Then the equation $f(x,y)=y^{3}-x = 0$ generates an implicit function $g:\mathbb{R}\rightarrow \mathbb{R}$ defined by $g(x)=x^{1/3}$

However, the derivative $g'(x)=\frac{1}{3}x^{-2/3}$ is not defined at $x = 0$ because for small changes $dx$ and $dy$ we have

$df(x,y) = f_{x}(x,y)dx + f_{y}(x,y)dy = 0$


$g'(x)=dy/dx=- f_{x}(x,y)dx / f_{y}(x,y)dy$

which is defined only when

$f_{y}(x,y)dy \neq 0$

  • $\begingroup$ Could you please consider formatting your answer. Math.SE uses MathJax to display mathematical expression. I believe you have misttook the syntax. It should be in the form $ <<Tex Commnad here>>$ $\endgroup$ – funktor Apr 11 '12 at 13:34
  • $\begingroup$ sorry, switching between SE and the browser LaTeX plugin I sometimes forget which one uses which syntax :D $\endgroup$ – scibuff Apr 11 '12 at 13:39
  • $\begingroup$ I believe Srijan is looking for a function defined only implicitly that is non-differentiable at every point in an open interval, not just at an isolated point. See en.wikipedia.org/wiki/Weierstrass_function $\endgroup$ – Barry Smith Apr 11 '12 at 13:48
  • $\begingroup$ right ... how about the Koch snowflake or space-filling curve $\endgroup$ – scibuff Apr 11 '12 at 14:16
  • $\begingroup$ @BarrySmith Your particular link seems to answer my question. Are there other such examples? $\endgroup$ – Srijan Apr 11 '12 at 16:16

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