Use the definition of the derivative to show that the function is not differentiable at the given value: (x-1)^(2/5) at x=1 I've been trying to figure out this problem for a while with no success. My main hurdle is figuring out the derivative of the function y=(x-1)^(2/5) using the definition of the derivative. At first I tried multiplying by a conjugate with a power of 4/5 but that did not work. I get the feeling I must use the (a^5-b^5) expansion, but I am not sure where to proceed from there. Thank you in advance.
Define the function $f: \mathbb{R} \mapsto \mathbb{R}$ by

$f(x)=(x-1)^{2/5}, \, \, \, x\in \mathbb{R}$.

We say that a function $f$ is differentiable in a point $x_0$ if the limit

$\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}$,

exists in $\mathbb{R}$. Now with $x_0=1$,

$\frac{f(x_0+h)-f(x_0)}{h} = \frac{(1-1+h)^{2/5}-(1-1)^{2/5}}{h}
=\frac{h^{2/5}}{h}=h^{-3/5}$.

As $h\to 0$ from above, this limit converges to $+\infty$. Thus $f$ is not differentiable in $x_0=1$.
Define the function $f: \mathbb{R} \mapsto \mathbb{R}$ by

$f(x)=(x-1)^{2/5}, \, \, \, x\in \mathbb{R}$.

We say that a function $f$ is differentiable in a point $x_0$ if the limit

$\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}$,

exists in $\mathbb{R}$. Now with $x_0=1$,

$\frac{f(x_0+h)-f(x_0)}{h} = \frac{(1-1+h)^{2/5}-(1-1)^{2/5}}{h}
=\frac{h^{2/5}}{h}=h^{-3/5}$.

As $h\to 0$ from above, this limit converges to $+\infty$. Thus $f$ is not differentiable in $x_0=1$.