Show that f is not differentiable at 0

Hey guys, I need help with this question that has been bugging me...

Let f(x) = x^{3/5}. Show that f is not differentiable at zero. Show that y = x^{3/5} has a vertical tangent line at (0,0)

I don't know how to prove the statements... Any tips and pointers (plus the method) will help alot...

Thanks!

Re: Show that f is not differentiable at 0

There are several **different** ways to do this, depending on what you can or want to use.

Of course, a function of one variable **is** "differentiable" at x= a if and only if $\displaystyle \lim_{x\to h} \frac{f(a+h)- f(a)}{h}$ exists. It is NOT differentiable if that limit does not exist. So you could show that $\displaystyle \lim_{h\to 0}\frac{h^{3/5}- 0}{h}$ does not exist.

Or you could use the fact that, while the derivative of a function is not necessarily continuous, it **does** have the "intermediate value property". In particular, if the derivative at x= a exists, then $\displaystyle \lim_{x\to a}f'(x)= a$. So what **is** the derivative of $\displaystyle f(x)= x^{3/5}$? What is the limit of that derivative as x goes to 0?