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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.

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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

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

Thus

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

which is defined only when

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

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    @scibuff That is i$n$deed a great resource.2012-04-11