Why does the cube root function have a discontinuous derivative at x=0?

[{???}]

Hey everyone,

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

Now, granted, the discontinuity in $f'(x)$ is a vertical asymptote where $\delta(M)$ can always be chosen such that:
$$|x|<\delta \Rightarrow f'(x)>M$$
So unlike the asymptotes of $\frac{1}{x}$ for example, or the derivative of $y=x^{2/3}$, the function both sides of the discontinuity "goes to positive infinity" so the tangent line to $x^{1/3}$ approaches the same slope on either side of $x=0$.

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)

 

[haruspex]

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)

Yes to the first, but I don't understand your reasoning in the second. How do you think it would lead to such? Mathematically smooth is very much more demanding than apparently smooth.

 

[{???}]

Yes to the first, but I don't understand your reasoning in the second. How do you think it would lead to such? Mathematically smooth is very much more demanding than apparently smooth.

My reasoning is best described mathematically as follows (and I admit, rather poorly at that):
Suppose we rotate the coordinate axes by $\frac{\pi}{6}$ (i.e., counterclockwise) and label the new grid $(x^*,y^*)$. We could technically have in this new coordinate system a function $y^*=f^*(x)$ that, although not expressible in terms of elementary functions, describes the graph of $y=x^{1/3}$.

My claim is that
$$\frac{\mathrm{d}y^*}{\mathrm{d}x}$$
is continuous. For other functions with "clearly" discontinuous derivatives, i.e., $y_2=|x|, y_3=x^{2/3},\ldots$ this is not the case.

Just wanted to point this out because it's different from most other piecewise-smooth functions in that respect. Any thoughts?

 

[kisengue]

Your question is kind of a non-question - your f(x) is not mathematically smooth, that is, it is not in C^inf. It is only in C^0 or D^1.

 

[haruspex]

Suppose we rotate the coordinate axes

OK, so you're thinking of it as a smooth curve in the plane, in more of a co-ordinate free sense.
That's true, but that is rather different from a function. E.g. you could rotate it so that there are multiple y values for some x values, so now it is not even a function in the standard definition.
One way around this is to parameterise, making x and y functions of some other variable that changes monotonically along the curve.