<s>[INST] How do I choose between $\lim_{x\to a} \frac {f(x) - f(a)}{x-a}\ $ and $\lim_{x\to a} \frac{f(a+h)-f(a)}{h}$? How do I choose between $\lim_{x\to a} \color{red}{\frac {f(x) - f(a)}{x-a}}$ and $\lim_{x\to a} \color{blue}{\frac{f(a+h)-f(a)}{h}}$?
Is it when I have more than two powers? For example, can I use the former for $\ y=3+4x^2-2x^3$?
 [/INST] How does one show that $\dfrac d {dx} x^n = nx^{n-1}$?  The way I first learned it was this:
\begin{align}
\frac d {dx} x^n & = \lim_{h\to 0}\frac{(x+h)^n-x^n}{h} \\[8pt]
& = \lim_{h\to0}\frac{\left( x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2+\frac{n(n-1)(n-2)}6x^{n-3}h^3+\cdots+h^n \right)-x^n}{h} \\[8pt]
& \text{etc.}
\end{align}
This is perfectly comprehensible, and I was amazed that it was possible to find the exact slope, but it is overkill.  Let's apply it to a concrete case:
\begin{align}
\frac d {dx} x^4 & = \lim_{h\to0} \frac{(x+h)^4-x^4}h \\[8pt]
& = \lim_{h\to0} \frac{(x^4+4x^3h+6x^2h^2+4xh^3+h^4)-x^4}h \qquad\longleftarrow\text{overkill} \\[8pt]
& = \lim_{h\to0} \frac{4x^3h+6x^2h^2+4xh^3+h^4}h \\[8pt]
& = \lim_{h\to0} (4x^3+6x^2h+4xh^2+ h^3) \\[8pt]
& = 4x^3 + 0 + 0 + 0.
\end{align}
Now here it is the other way:
\begin{align}
\frac d {dx} x^4 & = \lim_{w\to x}\frac{w^4-x^4}{w-x} \\[8pt]
& = \lim_{w\to x}\frac{(w-x)(w^3+w^2 x+wx^2 + x^3)}{w-x} \qquad\longleftarrow\text{not overkill} \\[8pt]
& = \lim_{w\to x} (w^3+w^2x+wx^2+x^3) \\[8pt]
& = x^3+x^3+x^3+x^3.
\end{align}
The reason the first version is overkill is that it requires you to use every binomial coefficient when you actually only need the first two.
Imagine this in the case of $x^{20}$
$$
(x^{20} + 20x^{19}h + 190 x^{18}h^2 + 1140 x^{17}h^3 + 4845 x^{16}h^4 + \cdots+h^{20})-x^{20}
$$
(and then $x^{20}$ cancels and then $h$ cancels)
versus
$$
(w-x)(w^{19}+w^{18}x+w^{17}x^2+w^{16}x^3+\cdots+x^{19})
$$
(and then $w-x$ cancels).
In the first version we have all those numbers:
$$
1, 20, 190, 1140, 4845, \ldots
$$
but only the first two are really needed.  Even showing that there are some such numbers, without evaluating them, is immense overkill for this problem.
Let's apply this to show $\dfrac d{dx}\sin x = \cos x$:
\begin{align}
\frac d {dx} \sin x & = \lim_{w\to x}\frac{\sin w - \sin x}{w-x} \\[8pt]
& = \lim_{w\to x}\frac{2\cos\left(\frac{w+x}2\right)\sin\left(\frac{w-x}2\right)}{w-x} \text{ (a trigonometric identity)} \\[8pt]
& = \left( \lim_{w\to x} \cos\left(\frac{w+x}2 \right)\right)\left( \lim_{w\to x} \frac{\sin\left(\frac{w-x}2\right)}{\left(\frac{w-x}2\right)} \right) \\[8pt]
& = \left( \lim_{w\to x} \cos\left(\frac{w+x}2 \right)\right)\left( \lim_{v\to0}\frac{\sin v} v \right) \\[8pt]
& = (\cos x)\cdot 1.
\end{align}
In this argument, only one basic limit, $\displaystyle\lim_{v\to0}\frac{\sin v}v=1$, is needed, whereas the more conventional argument needs two: that one and also that $\displaystyle\lim_{v\to0}\frac{1-\cos v}{v}=0$.
</s>