Find the derivative of $\sqrt[n]{x}$ using the formal definition of a derivative Given $\sqrt[n]{x}$, prove using the formal definition of a derivative that :
$$\frac{d}{dx} (\sqrt[n]{x}) = \frac{x^{\frac{1-n}{n}}}{n}$$
Now this would be ridiculously easy to show using the Power Rule, but alas, that is not the goal of this question.

Using the formal definition of a limit we get :
\begin{equation} 
\begin{split}
f'(x) & = \lim_{h \ \to \  0} \frac{f(x+h)-f(x)}{h}  \\
 & = \lim_{h \ \to \  0} \frac{\sqrt[n]{x+h}-\sqrt[n]{x}}{h} \\
 & = \lim_{h \ \to \  0} \frac{(x+h)^{\frac{1}{n}}-(x)^{\frac{1}{n}}}{h}
\end{split}
\end{equation}
But it is unclear to me how to proceed next, essentially all we need to do to get this limit into a determinate form (it currently is in an indeterminate form) is to factor out a $h$ in the numerator, but there doesn't seem to be an obvious way to do so.
What algebraic technique, would you use to factor out a $h$ in the numerator in this case? For $n=2$, you could easily multiply the fraction by the conjugate to get the limit into a determinate form, and for $n=3$, you could do the same with the help of a few identities, but how would you go about this for the general case, as stated in the example I've given above.
This question is the general $n^{th}$ case of finding the derivative using the formal definition, for functions such as $f(x) = \sqrt{x}$, $f(x) = \sqrt[3]{x}$ and so forth, and is aimed at finding the best algebraic technique to manipulate the limit to get it into a determinate form.
 A: Using the identity
$$ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \dots + ab^{n-2} + b^{n-1}) $$
you have
$$ \frac{(x+h)^{\frac{1}{n}} - x^{\frac{1}{n}}}{h} \cdot \frac{(x+h)^{\frac{n-1}{n}} + (x+h)^{\frac{n-2}{n}}x+\dots+(x+h)^{\frac{1}{n}}x^{\frac{n-2}{n}}+x^{\frac{n-1}{n}}}{(x+h)^{\frac{n-1}{n}} + (x+h)^{\frac{n-2}{n}}x+\dots+(x+h)^{\frac{1}{n}}x^{\frac{n-2}{n}}+x^{\frac{n-1}{n}}} = \frac{1}{(x+h)^{\frac{n-1}{n}} + (x+h)^{\frac{n-2}{n}}x+\dots+(x+h)^{\frac{1}{n}}x^{\frac{n-2}{n}}+x^{\frac{n-1}{n}}} \xrightarrow[h \to 0]{} \frac{1}{x^{1-\frac{1}{n}}+x^{1-\frac{2}{n}}x^{\frac{1}{n}}+\dots +x^{\frac{1}{n}}x^{1-\frac{2}{n}}+x^{1-\frac{1}{n}}}=\frac{1}{n x^{1 - \frac{1}{n}}}=\frac{x^{\frac{1}{n}-1}}{n}. $$
A: Multiply top and bottom by
$$\sum_\limits{k=0}^{n-1} (x+h)^\frac{k}{n}(x)^\frac{n-1-k}{n}$$
this gives you.
$$\lim_\limits{h\to 0} \dfrac {(x+h)-x}{h\sum_\limits{k=0}^{n-1} (x+h)^\frac{k}{n}(x)^\frac{n-1-k}{n}}$$
And evaluate as $h$ goes to $0.$
A: The question is definitely not trivial. +1 for OP. The solution follows from the following theorem:
Theorem: If $a > 0$ and $n$ is a rational number then $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}\tag{1}$$ This is one of the standard limits which can be used to evaluate many limits involving algebraic functions.
The proof of the above theorem is easy if $n$ is an integer. For positive integers we can simply use $$x^{n} - a^{n} = (x - a)\sum_{i = 0}^{n - 1}x^{n - 1 - i}a^{i}$$ For $n = 0$ the result is obvious. For negative integer $n = -m$ we can use $x^{n} = 1/x^{m}$ and the fact that the result holds for positive integers. Similarly if the result holds positive rational number $n$ we can show that it holds for negative rational $n$ also.
Thus we need to show that if $n = p/q$ with integers $p > 0, q > 1$ then the formula $(1)$ holds. Let $b = a^{1/q}$ so that $a = b^{q}$. We know that $$\lim_{y \to b}\frac{y^{q} - b^{q}}{y - b} = qb^{q - 1}\tag{2}$$ From $(2)$ it follows that the ratio $(y^{q} - b^{q})/(y - b)$ is bounded and away from $0$ as $y \to b$. Hence its reciprocal is also bounded and away from $0$ as $y \to b$. Also note that when $y \to b$ then $x = y^{q} \to b^{q} = a$ (and vice-versa because $f(y) = y^{q}$ is strictly monotone in $[0, \infty)$). Thus the ratio $(x^{1/q} - a^{1/q})/(x - a)$ is bounded when $x = y^{q} \to a$ and therefore $$\lim_{x \to a}x^{1/q} = a^{1/q}\tag{3}$$ (this by the way proves continuity of $x^{1/q}$).
Now we have
\begin{align}
L &= \lim_{x \to a}\frac{x^{n} - a^{n}}{x - a}\notag\\
&= \lim_{x \to a}\frac{x^{p/q} - a^{p/q}}{x - a}\notag\\
&= \lim_{t \to b}\frac{t^{p} - b^{p}}{t^{q} - b^{q}}\text{ (putting }x = t^{q}, a = b^{q}\text{ and using (3))}\notag\\
&= \lim_{t \to b}\dfrac{\dfrac{t^{p} - b^{p}}{t - b}}{\dfrac{t^{q} - b^{q}}{t - b}}\notag\\
&= \frac{pb^{p - 1}}{qb^{q - 1}}\notag\\
&= \frac{p}{q}b^{p - q}\notag\\
&= na^{n - 1}\notag
\end{align}
There is another way to prove this (via inequalities and squeeze theorem) without using the continuity of $x^{1/q}$. Let me know if you are interested in that version.

Update: On request of OP I am providing a proof of formula $(1)$ based on Squeeze Theorem. The credit for this proof must go to G. H. Hardy!
In what follows all the numbers are positive (whether they are integers, rationals or reals will be mentioned as and when needed).
Let $a, b$ be real numbers with $a > 1 > b > 0$. Let $r$ be an integer. Clearly we have $a^{r} > a^{i}$ for all $i = 0, 1, 2, \ldots, r - 1$. Hence on adding these inequalities we get $$ra^{r} > 1 + a + a^{2} + \cdots + a^{r - 1}$$ Multiplying by $(a - 1) > 0$ we get $$ra^{r}(a - 1) > a^{r} - 1$$ Adding $r(a^{r} - 1)$ on both sides, and dividing by $r(r + 1)$, we obtain $$\frac{a^{r + 1} - 1}{r + 1} > \frac{a^{r} - 1}{r}\tag{4}$$ Similarly we can prove that $$\frac{1 - b^{r + 1}}{r + 1} < \frac{1 - b^{r}}{r}\tag{5}$$ It follows that if $r, s$ are positive integers with $r > s$ then $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s},\,\frac{1 - b^{r}}{r} < \frac{1 - b^{s}}{s}\tag{6}$$ If we put $s = 1$ we get $$a^{r} - 1 > r(a - 1),\, 1 - b^{r} < r(1 - b)\tag{7}$$ for $r > 1$.
Next we show that the inequalities $(6), (7)$ hold when $r, s$ are positive rational numbers with $r > s$. Let $r = k/l, s = m/n$ and $r > s$ implies that $kn > lm$. Let $c = a^{1/ln}$ so that $c > 1$. In the first inequality of $(6)$ we can replace $a$ by $c$, $r$ by $kn$ and $s$ by $lm$ to get $$\frac{c^{kn} - 1}{kn} > \frac{c^{lm} - 1}{lm}$$ or $$\frac{a^{r} - 1}{r} > \frac{a^{s} - 1}{s}$$ In similar manner we can prove that other inequalities also hold when $r, s$ are rational numbers. Now that $r, s$ are rational, it is possible to take $r = 1$ in $(6)$ to get $$a^{s} - 1 < s(a - 1),\,1 - b^{s} > s(1 - b)\tag{8}$$ for rational $s$ with $0 < s < 1$. Thus we have inequalities $(6)-(8)$ for all positive rational numbers $r, s$ with $r > 1 > s$.
In what follows we will assume that $a, b$ are real with $a > 1 > b > 0$ (same as before) and $r, s$ are rational with $r > 1 > s > 0$. Clearly $1/b > 1$ and hence replacing $a$ by $1/b$ and $b$ by $1/a$ in $(7)$ we get $$a^{r} - 1 < ra^{r - 1}(a - 1),\, 1 - b^{r} > rb^{r - 1}(1 - b)\tag{9}$$ Similarly from $(8)$ we get $$a^{s} - 1 > sa^{s - 1}(a - 1),\, 1 - b^{s} < sb^{s - 1}(1 - b)\tag{10}$$ Combining $(7)$ and $(9)$ we get $$ra^{r - 1}(a - 1) > a^{r} - 1 > r(a - 1)\tag{11}$$ Writing $a = x/y$ we get $$rx^{r - 1}(x - y) > x^{r} - y^{r} > ry^{r - 1}(x - y)\tag{12}$$ for $x > y > 0$. Similarly from $(8)$ and $(10)$ we get $$sx^{s - 1}(x - y) < x^{s} - y^{s} < sy^{s - 1}(x - y)\tag{13}$$ for $x > y > 0$.
From the above inequalities it is clear that the function $f(x) = x^{r}$ is continuous for $x > 0$. Taking reciprocals it is easy to see that the function $f(x)$ is continuous even if $r$ is negative rational number. Further if we divide by $(x - y) > 0$ and let $x \to y^{+}$ we get via Squeeze Theorem the fundamental result $$\lim_{x \to y^{+}}\frac{x^{r} - y^{r}}{x - y} = ry^{r - 1}$$ for all positive rational numbers $r$ and $y > 0$. Interchanging the roles of $x, y$ it is easy to see that the limit holds for $x \to y^{-}$. This proves the formula $(1)$ for positive rational values of $n$.
This is the way Hardy proves the formula $$\frac{d}{dx}(x^{n}) = nx^{n - 1}$$ for rational $n$ in his classic text "A Course of Pure Mathematics".
A: $$\frac{d}{dx}x^n=\lim_{h\to0}\frac{(x+h)^n-x^n}h$$
$$=\lim_{h\to0}\frac{[x^n+nx^{n-1}h+\frac{n(n-1)x^{n-2}h^2}{2!}+\dots]-x^n}h$$
named Binomial Expansion.
$$=\lim_{h\to0}\frac{{nx^{n-1}h+\frac{n(n-1)x^{n-2}}{2!}h^2+\dots}}{h}$$
$$=\lim_{h\to0}nx^{n-1}+\frac{n(n-1)x^{n-2}}{2!}h+\dots$$
$$=nx^{n-1}+0+0+\dots=nx^{n-1}$$
Just use $n=\frac1m$ for your case.
A: If $f(x) =x^{1/n}$ then $f(x)^n=x$. Now derive using the chain rule and solve the result for $f'$. 
If you really need to use the limit of a difference quotient, note that he above reasoning is really the same as doing the following:
Write
$$
f'(x) = \lim_{y\to x}\frac{\sqrt[n]{y}-\sqrt[n]{x}}{y-x}
$$
with the substitution $t = \sqrt[n]{x}$, $s=\sqrt[n]{y}$ which then gives
$$
f'(x) = \lim_{s\to t}\frac{s-t}{s^n-t^n} = \left(\lim_{s\to t}\frac{s^n-t^n}{s-t}\right)^{-1} = (nt^{n-1})^{-1} = \frac{1}{n}x^{-\tfrac{n}{n-1}}
$$
(where we used that the derivative of $t^n$ is $nt^{n-1}$).
