Derivative result is $0 \over 0$. Does it imply the point isn't differentiable? $$f'(x) = {2x \over {4(x^2)^{3 \over 4}}}$$
for $x_0$ the derivative is $0 \over 0$, Which is defined as infinity as much as I know.
Is it sufficient to say the above in order to prove $x_0=0$ isn't differentiable, or should I explain it in another way?  
Thanks
 A: Two things:
(1) Canonically, $\frac00$ has no definition — which is different from being defined as infinity.  In fact, the classic definition of a derivative involves the limit of two things which go to $0$: just think of, for instance, taking the derivative of $f(t)=t^2$, where the result is $f'(x) = \lim\limits_{h\to0}\dfrac{(x+h)^2-x^2}{h}=\lim\limits_{h\to0}\dfrac{2xh+h^2}{h}$.  Clearly this is a $\frac00$ limit, but that doesn't mean that the derivative is necessarily infinite.
(2) You can simplify your expression substantially: rewrite the denominator as a direct power of $x$ using the rule $\left(a^b\right)^c=a^{bc}$, then rewrite the entire fraction as a single term in $x$ using the rule $\frac{a^b}{a^c}=a^{b-c}$.  You should be able to clean up the constant terms in the process.  Once you do this, you'll find that you no longer have a $\frac00$ limit and can evaluate it more directly.
A: The expression "$\frac{0}{0}$" is not infinity, but rather it is undefined. When you get something like that, you should check using limits what happens. You have to compute both
$$\lim_{x\to0^+}f'(x)$$
$$\lim_{x\to0^-}f'(x)$$
If they disagree, or are infinite, then $f'$ is not continuous at $0$, if they agree to some finite value, then that value is $f'(0)$.
