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I was just curious about the nature of the cube root function [itex]f(x)=x^{1/3}[/itex]. I know that its derivative is obviously [itex]\frac{1}{3}x^{-2/3}[/itex] which has a discontinuity at [itex]x=0[/itex]. However, in the non-mathematical sense, the graph of [itex]y=f(x)[/itex] looks smooth - I don't see any angles or cusps like [itex]y=|x|[/itex] or [itex]y=x^{2/3}[/itex].

Now, granted, the discontinuity in [itex]f'(x)[/itex] is a vertical asymptote where [itex]\delta(M)[/itex] can always be chosen such that:

[tex]|x|<\delta \Rightarrow f'(x)>M[/tex]

So unlike the asymptotes of [itex]\frac{1}{x}[/itex] for example, or the derivative of [itex]y=x^{2/3}[/itex], the function both sides of the discontinuity "goes to positive infinity" so the tangent line to [itex]x^{1/3}[/itex] approaches the same slope on either side of [itex]x=0[/itex].

My question is: Is this a good enough explanation as to why an apparently smooth function has a discontinuous derivative? If not, why doesn't this violate the definition of a (mathematically) smooth function? (This may border on a philosophical question, so my apologies in advance if so)

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# Continuity of d/dx [x^(1/3)]

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