TITLE: What's really going on behind calculus?
QUESTION [26 upvotes]: I'm currently taking maths at A Level and I have found it strange that it is not explained why the differential of $x^n$ is $nx^{n-1}$, for example. I can see that it works through observation and first principle, but how can it be derived? And what about for an unknown function, not based on trigonometry, $e^x$ or polynomials? Is there some sort of intuition or derivation that can lead to a general answer, other than making observations? Another thing that I do not quite understand is why higher derivatives of some function is denoted by $\frac{d^ny}{dx^n}$.
Sorry if this sounds really basic but I would appreciate any explanations/links to good resources (I've had a look online and couldn't find too much other than some nice explanations special cases etc). Thanks :)

REPLY [4 votes]: Concerning your question of why higher derivatives are denoted by $$\frac{d^ny}{dx^n}$$
There is a reason for this notation. 
Now $$\begin{align}\frac{d^2y}{dx^2} &= \lim_{h \to 0} \frac{y'(x + h) - y'(x)}{h}\\&=\lim_{h \to 0}\frac{\lim_{h_1 \to 0} \frac{y(x + h + h_1) - y(x + h)}{h_1} - \lim_{h_2 \to 0} \frac{y(x + h_2) - y(x)}{h_2}}h\\&=\lim_{h \to 0}\lim_{h_1 \to 0}\lim_{h_2 \to 0} \frac{ \frac{y(x + h + h_1) - y(x + h)}{h_1} - \frac{y(x + h_2) - y(x)}{h_2}}h\end{align}$$
Now assuming both the iterated and the combined limit exist, we can set the three values $h, h_1, h_2$ to be equal without changing the value. So
$$\begin{align}\frac{d^2y}{dx^2} &=\lim_{h \to 0} \frac{ \frac{y(x + 2h) - y(x + h)}{h} - \frac{y(x + h) - y(x)}{h}}h\\&=\lim_{h \to 0} \frac{ y(x + 2h) - 2y(x + h) + y(x)}{h^2}\end{align}$$
The expression on the top is the "2nd difference". It is what you get from taking a difference of a difference, so it is the taking of differences that is squared in the short-hand notation: $d^2y$. But the denominator is just $h^2$, which is the square of a single difference in $x$. Thus $dx^2$ (which is intended to mean $(dx)^2$, not $d(x^2)$).