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

1. Jul 9, 2012

### {???}

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)

2. Jul 10, 2012

### haruspex

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.

3. Jul 10, 2012

### {???}

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?

4. Jul 10, 2012

### 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.

5. Jul 10, 2012

### haruspex

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.