# Finding the derivative using the definition

show that f(x)= x^(1/3) is not differentiable at zero.
I know there is a vertical tangent at the origin and therefor not differentiable, but I'm not sure how to say it in analysis.

I have tried using caratheodry but it doesn't seem to get me anywhere

Can you find the derivative of f in 0?? Perhaps by applying the definition of derivative:

$$\lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}$$

What is that derivative? Does it exist?

No , I guess I should have said show show not differentiable using the definition.

Anyways

if I take the derivative I get

1/3 x^(-2/3)

if x goes to 0 is undefined... But how do you say that in math talk

You just say that the derivative is undefined. If you calculate the limit from the definition, then you say that the limit does not exist (because...?). When you're arguing like in your last post, you just say that it is undefined in 0 as division by 0 is undefined.

does it violate the Archimedian property? I'm not sure how to use the definition because how do you expand (x+h)^(1/3)-(x)^1/3. I cant use the binomial theorem to expand it out because that only works for whole numbers.

does it violate the Archimedian property? I'm not sure how to use the definition because how do you expand (x+h)^(1/3)-(x)^1/3. I cant use the binomial theorem to expand it out because that only works for whole numbers.

Do note that you take the limit in 0, so you can take x=0. This simplifies a whole lot.