Cube Root function not differentiable Why is the cube root function not differentiable at $x=0?$
I graphed it and the curve looks a bit vertical at that point, is that why? Can someone give a good explanation please.
 A: Suppose a function $f(x)=x^n$; its derivative is $f'(x)=nx^{n-1}$. If $n<1$, then rewrite $$f'(x)=nx^{n-1}=\frac{n}{x^{1-n}}$$ the exponent being positive in the denominator, $f'(x)$ is undefined for $x=0$. So, the curve is not a bit vertical.
For you specific case of $f(x)=x^{1/3}$, $f'(x)=\frac{1}{3 x^{2/3}}$, compute the value of $f'(10^{-n})$ for $n=10$ and $n=100$. 
A: Yes, the easiest explanation is that the tangent line is vertical at $x=0$ and a vertical line has an undefined slope. However, this misses the part that is a bit paradoxical.
If the slope of the tangent line is vertical, the graph of the function is vertical at that instant. But if the graph of the function is vertical, it can never "get past" $x=0$. How do you ever get to $x=0.1$, if the function is going straight up at $x=0$?
A: You can derive this formally, as well, using the $\epsilon-\delta$ definitions of limits.
Remember that the definition of the derivative of a function $f$ at location $a$ is:
$f'(a):= \displaystyle \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$
Applying the definition when $f(x)=\sqrt[3]{x}$ and $a=0$, we are interested in determining the following limit: $\displaystyle \lim_{h \to 0} \frac{\sqrt[3]{0+h}-\sqrt[3]{0}}{h}=\displaystyle \lim_{h \to 0} \frac{\sqrt[3]{h}}{h}$
Claim:

$\displaystyle \lim_{h \to 0}\frac{\sqrt[3]{h}}{h}=\infty$


In order to prove this, we need to show that for $L \gt 0$, we can construct a $\delta \gt 0$ such that for any $x$ in $\mathbb R$, we have the following:
$$0 \lt |h| \lt \delta \rightarrow \frac{\sqrt[3]{h}}{h}\gt L$$
Before constructing $\delta$, note that $ \frac{\sqrt[3]{h}}{h}$ can be rewritten using the rules of exponents as $\frac{1}{\sqrt[3]{h^2}}$.
First, consider when $h \lt 0$. Let $h  \gt \frac{-1}{\sqrt{L^3}}$. From this we can deduce the following:
$$0 \gt h \gt \frac{-1}{\sqrt{L^3}} \implies 0 \lt -h \lt \frac{1}{\sqrt{L^3}}$$
Note that because $h \lt 0$, $\sqrt{h^2}=-h$. Therefore, we can rewrite our above inequality as:
$$ 0 \lt \sqrt{h^2} \lt \frac{1}{\sqrt{L^3}} \implies 0 \lt h^2 \lt \frac{1}{L^3}\implies 0 \lt \sqrt[3]{h^2}\lt \frac{1}{L} \implies 0 \lt L \lt \frac{1}{\sqrt[3]{h^2}} $$
Next, consider when $h \gt 0$. Let $h \lt \frac{1}{\sqrt{L^3}}$. Note that because $h$ is positive, $\sqrt{h^2}=h$. Then we can write:
$$0 \lt h \lt \frac{1}{\sqrt{L^3}} \implies 0 \lt \sqrt{h^2} \lt \frac{1}{\sqrt{L^3}} \cdots \text{ same steps as before } \cdots \implies 0 \lt L \lt \frac{1}{\sqrt[3]{h^2}}$$.
From these two results, we can let our $\delta = \frac{1}{\sqrt{L^3}}$. i.e. for $h \neq 0$, as long as $\frac{-1}{\sqrt{L^3}} \lt h \lt \frac{1}{\sqrt{L^3}}$, we can guarantee that $\frac{\sqrt[3]{h}}{h}=\frac{1}{\sqrt[3]{h^2}} \gt L$, as desired.
