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 $\ y=3+4x^2-2x^3$?
 A: They are equivalent definitions: We know 
$$ f'(a) = \lim_{x \to a } \frac{ f(x) - f(a) }{x-a} $$
Now, put $h = x -a $. Notice $x \to a \iff x - a \to 0 \iff h \to 0 $. Hence, 
$$ f'(a) = \lim_{h \to 0 } \frac{ f(h+a) - f(a) }{h } $$
A: Note that
$$
\lim_{x\to a} \frac{f(a+h)-f(a)}{h}
$$
means just
$$
\frac{f(a+h)-f(a)}{h}
$$
because there's no $x$ in the fraction. You probably mean
$$
\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}
$$
which is exactly the same as
$$
\lim_{x\to a} \frac{f(x)-f(a)}{x-a}
$$
The former is usually more manageable, because you don't have to collect $x-a$, but simply $h$.
If $f(x)=3+4x^2-2x^3$, then $f(a)=3+4a^2-2a^3$ and
$$
f(a+h)=3+4(a+h)^2-2(a+h)^3=3+4a^2+8ah+4h^2-2a^3-6a^2h-6ah^2-2h^3
$$
so
$$
f(a+h)-f(a)=8ah+4h^2-6a^2h-6ah^2-2h^3=h(8a+4h-6a^2-6ah-2h^2)
$$
and therefore
$$
\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=
\lim_{h\to0}(8a+4h-6a^2-6ah-2h^2)=8a-6a^2
$$
With $(f(x)-f(a))/(x-a)$ the computation is less obvious.

The method actually can be greatly simplified when polynomials are concerned, because of a nice property: when you expand
$$
(a+h)^n
$$
you can simply write it as
$$
a^n+na^{n-1}h+h^2P_n(a,h)
$$
where $P_n(a,h)$ is a polynomial expression in $a$ and $h$. Why is this? The case $n=1$ is obvious; then
\begin{align}
(a+h)^{n+1}&=(a+h)^n(a+h)\\
&=(a^n+na^{n-1}h+h^2P_n(a,h))(a+h)\\
&=a^{n+1}+na^{n}h+ah^2P_n(a,h)+a^nh+na^{n-1}h^2+h^3P_n(a,h)\\
&=a^{n+1}+(n+1)a^nh+h^2\bigl(aP_n(a,h)+na^{n-1}+hP_n(a,h)\bigr)
\end{align}
and we have proved the thesis, because certainly
$$
P_{n+1}=aP_n(a,h)+na^{n-1}+hP_n(a,h)
$$
is a polynomial expression in $a$ and $h$ once $P_n(a,h)$ is.
So, in the case of $f(x)=x^{20}$ we just need
\begin{align}
\lim_{h\to0}\frac{(a+h)^{20}-a^{20}}{h}
&=\lim_{h\to0}\frac{a^{20}+20a^{19}h+h^2P_{20}(a,h)-a^{20}}{h}\\
&=\lim_{h\to0}\bigl(20a^{19}+hP_{20}(a,h)\bigr)\\
&=20a^{19}
\end{align}
You don't need the full development as it would be in the case of $x^{20}-a^{20}$.
A: This is a good question. Both of these formulas will give you the VALUE of the derivative of a function at the particular point $a$. 
When we say $a$, it means we are plugging in the value $a$ into our function before we do our calculations. 
Compare these to the formula $$f^{\prime}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$ this expression gives you the derivative FUNCTION. 
So, the first two formulas give the same value of the derivative at the point $a$ and the formulas $f'(x)$, known as the definition of the derivative, gives the derivative function for a differentiable function.
A: 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 $\displaystyle\lim_{v\to0}\frac{1-\cos v}{v}=0$.
