Using dimensional analysis to evaluate $\frac{d}{dx}x^n$ Let $x$ have dimensions $[L]$ of length, so that $dx$ also has dimension $[L]$. Then $$\frac{d(x^n)}{dx}\;\text{has dimension}\;\frac{[L]^n}{[L]}=[L]^{n-1}.$$ Therefore $$\frac{d}{dx}x^n=cx^{n-1}$$ for some (dimensionless) constant $c$. 
Disregarding the fact that $c$ is undetermined (bonus: can anyone derive it using dimensional analysis?), I am wondering: is this proof valid?
 A: It's not quite a proof due to the fact that it gives no information about $c$, the only thing you can say is that 
$$ \frac{d}{dx} x^n = c(x) x^{n-1} $$
by unit analysis [an example is $c(x) = \sin( x/L)$ where $L$ has units of length]. In this form we have no idea what $c(x)$ should be, and really can't say anything about the function.
Edit: Note that $c(x)$ is unit-less.
A: Suppose a function $f$ has the property that $f(\alpha x)=\alpha^n f(x)$ for all $\alpha$. Then $f(x)=cx^n$ for some constant $c$.
Now take a plot the graph of $y=x^n$ and scale it by a factor of $\alpha$ in the horizontal direction and by a factor of $\alpha^n$ in the vertical direction. The above property means you end up with the same graph you started with.
Now imagine a tangent line drawn on the graph at the point $(x,x^n)$. Scaling graphs doesn't change the fact that a tangent line is a tangent. So the tangent becomes the tangent at $(\alpha x,\alpha^nx^n)$.
But what happens to the gradient of a line if you scale a graph? If you scale by $\alpha$ horizontally and $\beta$ vertically the gradient is scaled by $\beta/\alpha$. So the gradient is scaled by $\alpha^{n-1}$ in this case.
In other words, the gradient at $(\alpha x, \alpha^nx)$ is the same as the gradient at $(x, x^n)$ scaled by $\alpha^{n-1}$.
So the gradient of $f$, $f'$, has the property that $f'(\alpha x)=\alpha^{n-1}f'(x)$.
So $f'(x)=cx^{n-1}$.
But what does this have to do with what you said? Dimension arguments are arguments about scaling in disguise. You can see this if you imagine plotting a graph of some physical quantity using some units for the scales. If you switch to different units your graph will get scaled. But fundamentally, changing units doesn't really change anything, and this fact means that when you scale your graph there are still going to be features that remain unchanged. Describing this in full generality will require more space than I have here but the argument above is one specific example.
Here's a mathematical (but high level and tricky) discussion of this subject by Terry Tao: https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/
In other words, your argument is valid, but only if you've previously proved the properties you need from dimensional analysis.
A: It's not valid, for the reasons that TylerHG and Javier Badia already alluded to in their comments. For example we could use your reasoning to say that if $v$ has dimension $L$, and $c$ is some physical constant also with dimension $L$, then $\frac{d(\sqrt{c^2-v^2})}{dv}$ has no dimension and hence is some constant. The problem is that dimensional analysis is merely a heuristic and doesn't take constants into account. Moreover, it discards information about the function, and so your argument gives the right answer for only $x^n$. Lest you think this kind of function is contrived, it in fact appears in relativitistic formulae where $c$ is of course the speed of light.
[Edit: As requested, I'll try to specify more precisely what is wrong with the argument and what is needed to make it correct.]
One objection to the above counter-example might be that it has more than one variable or constant. But there is still another problem with the argument, which is that there is no a priori reason to suppose that $x^n$ is physically meaningful when $x$ has some dimension $L$, and ${L}$ is not a dependent set ($x$ is not dimensionless). Under both those assumptions, the only possible physically meaningful dimensionless functions of $x$ are the dimensionless constant functions, and hence the only possible physically meaningful functions of $x$ that have dimension $L^k$ are of the form $cx^k$ for some constant $c$. For this to apply to the original question you have to justify why you think $x^n$ is physically meaningful for each $n$, otherwise it is not a valid proof.
In case one might think that the assumptions stated above are obviously valid ones, one should consider say $x^\pi$ or $x^{1000}$. Are they really physically meaningful? If not, how can one use physical units to argue anything about the behaviour of the mathematical functions? If one thinks that the result should simply hold for any $n$ since it holds for some like $1,2,3$, then it's already a fallacy. And even if it holds for all rational $n$, there is still no reason to think that it holds for irrational $n$. All these properties come about because of the mathematical nature of such functions, not physical.
A: Consider the formal definition of the derivative:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}.$$
Suppose that we could apply dimensional analysis under this operation. Then,
$$f'(x) = \lim_{h \to 0} \frac{c_1 \operatorname{dim}(f(x+h)) - c_1\operatorname{dim}(f(x))}{c_2 \operatorname{dim}(h)}.$$
Dimensionally speaking, the relation $x\propto \mathcal{U} \implies x = ku$ holds for every constant $k$. Therefore, we could use this definition to show that the derivative is any ratio we desire: $\frac{c_1}{c_2}$. This is nonsensical.
The reason this fails is because dimensionality must be finite and nonzero. Zero pounds is the same as zero meters -- zero is zero; it is a dimensionally-annihilating quantity. $dx$ does not have dimensionality $\mathcal{L}$, since $dx$, depending on how we treat it, is smaller than any finite quantity. As such we cannot use its dimensionality in an argument; it is not zero, but we cannot say what it is. Therefore, we cannot dimensionally derive calculus using these arugments; besides, calculus must hold even in the case of dimensionless quantities!
A: Given
$$
\frac{d}{dx} x^n = c x^{n-1}.
$$
Write this as
$$
\frac{d}{dx} x^p = c(p) x^{p-1},
$$
thus
$$
\frac{d}{dx} x^{p+q} = c(p+q) x^{p+q-1}.
$$
But
$$
\frac{d}{dx} x^{p+q} = \Big( \frac{d}{dx} x^p \Big) x^q + x^p \Big( \frac{d}{dx} x^q \Big) = \Big( c(p) + c(q) \Big) x^{p+q-1},
$$
therefore
$$
c(p+q) = c(p) + c(q),
$$
so we obtain
$$
c(p) = c(1) p.
$$
We also have
$$
\frac{dx}{dx} = 1,
$$
so
$$
c(1) x^0 = 1,
$$
whence
$$
c(1) = 1.
$$
Therefore
$$
c(p) = p,
$$
and we obtain
$$
\frac{d}{dx} x^n = n x^{n-1}.
$$
