Proving $x^{1/5}$ is differentiable The complete question is to prove that $f: \mathbb{R} \to \mathbb{R}$ $f(x) = x^{1/5}$ is continuous everywhere and differentiable everywhere but at x = 0
so I figured I'd prove that it's differentiable everywhere and proving the rest is easy.
I've made two attempts, one of which is most certainly wrong, the other I don't know how to finish:
Firstly, in class we proved that $x^n$ is differentiable when x > 0 (using the exponent and logarithm). So I will just consider when x < 0
$x<0$ let $x = -m$ where $m > 0$
so we need to show that $ \lim_{h\to 0} \dfrac{(-m+h)^{1/5} + m^{1/5}}{h} $ exists
$ \dfrac{(-m+h)^{1/5} + m^{1/5}}{h} = \dfrac{\left(-m^{1/5} + \dfrac{h}{5m^{4/5}} -\dfrac{2h^2}{25m^{9/5}} + \cdots\right) + m^{1/5}}{h} = \dfrac{1}{5m^{4/5}} + \dfrac{2h}{25m^{9/5}} + \cdots$
so $ \lim_{h\to 0} \dfrac{(-m+h)^{1/5} + m^{1/5}}{h}  = \lim_{h\to0}\dfrac{1}{5m^{4/5}} + \dfrac{2h}{25m^{9/5}} + ... = \dfrac{1}{5m^{4/5}} = \dfrac{-1}{5x^{4/5}} $ so the limit exists so f is differentiable for x < 0 
well I realised later that this attempt is incorrect as the binomial expansion assumes differentiability, but I put here to show what I'm thinking so far
my second attempt:
by using the definition $ \lim_{x \to c} \dfrac{f(x) - f(c)}{x-c}  = \lim_{x \to c} \dfrac{x^{1/5} - c^{1/5}}{x-c} = \lim_{x \to c}\dfrac{(x-c)(x^{-4/5} + cx^{-9/5} + c^2x^{-14/5} +...)}{x-c} $
but the last series is an infinite sum no?, is there anyway I can evaluate this?
if anyone could help me progress, or show me another way it'd be appreciated - thank you.
attempted a response to one of the answers below, could anyone help further?
 A: Differentiability of $f(x)=x^{1/5}$ for $x\not=0$ is slickly demonstrated if you write the derivative as
$$f'(x)=\lim_{u\to x}{x^{1/5}-u^{1/5}\over x-u}$$
and apply the factorization $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$ with $a=x^{1/5}$ and $b=u^{1/5}$.  This gives a factorization of the denominator  $x-u=a^5-b^5$, leading to
$$f'(x)=\lim_{u\to x}{1\over x^{4/5}+x^{3/5}u^{1/5}+x^{2/5}u^{2/5}+x^{1/5}u^{3/5}+u^{4/5}}={1\over5x^{4/5}}$$
Note, though, the final step assumes continuity of the fifth-root function (and its powers).  So if you're trying to prove differentiability and continuity in one fell swoop, this doesn't quite do it.
A: For every $y \in \mathbb{R}$, there exists one and only one $x \in \mathbb{R}$ such that $x^5=y$, and you call $x=y^{1/5}$ or $\sqrt[5]{y}$. So your function is the inverse function of $x \mapsto x^5$, and you apply the rule for the differentiation of the inverse function at any $x \neq 0$.
Alternatively, you can remark that
$$
x^{1/5} = \exp \left( \frac{1}{5} \log x \right)
$$
at least for $x>0$. Now you can easily differentiate the right-hand side. If $x<0$, $x^{1/5} = - (-x)^{1/5}$.
A: Is implicit differentiation OK... $[y(x)]^5=x$.
