first term in the product rule At the moment I am studying a Calculus book. This book states that if you have a function defined like this:
$
 f(x)=x^3
$
Than if you expand that function according to the Power Rule with the Binomial Theorem the derivative would be $3x^2$. I do understand that the derivative is $3x^2$. However it does not make sense when I use that theorem and then simplify the it.
The simplification of the derivative:
$
 f(x)= \frac{ f(x + h)^3 - f(x)^3 }{ h }
$
$
 f(x)= \frac{  (x + h)(x + h)(x + h) - x^{3} }{ h }
$
$
 f(x)= \frac{  x^{2}+ 2xh + h^{2}(x + h) - x^{3} }{ h }
$
$
 f(x)= \frac{  x^{3} + x^{2}h + 2x^{2}h + 2xh^{2} + h^{2} + h^{3}- x^{3} }{ h }
$
$
 f(x)= 3x^{2}  + 3xh +  h^{2}
$
So the book states, that because of that every term, except the first one, has h as a factor and therefore approaches 0. Therefore the answer is.
$
 f(x)= 3x^{2} 
$
Why is it, that only the first term is important and not the other two?
$
 f(x)= 3x^{2}  + 3xh +  h^{2}
$
 A: First, it is not convenient to denote 
$$
 \frac{ f(x + h)^3 - f(x)^3 }{ h }
$$
by $f(x)$ for two reasons: we have already defined $f$ to be the function $x\mapsto x^3$ and the variable of this function is $h$ since finally we pass to the limit $h\to0$.
Now to answer your question: if we pass to the limit $h\to 0$ the terms $3xh$ and $h^2$ vanishes and remains the term $3x^2$ which represents the derivative of $x^3$.
A: The derivative of $f(x)$ is denoted by $f'(x)$ and is defined as
$\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}h$.
If you choose $f(x)=x^3$, we have
$f'(x)=\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}h=\lim_{h\rightarrow0} \dfrac{(x+h)^3-x^3}h=\lim_{h\rightarrow0}\dfrac{3x^2h+3xh^2+h^3}h$.
Intuitively, $\lim_{x\rightarrow a}g(x)$ is the expression to which the value of  the function $g$ gets close to when evaluated at values very close to $a$ but not at $a$.
Since for all $h\ne 0$, $\dfrac{3x^2h+2xh^2+h^3}h=3x^2+3xh+h^2$, the derivative is given by $\lim_{h\rightarrow 0}3x^2+3xh+h^2$.
Now as $h$ approaches $0$, $3x^2+3xh+h^2$ gets close to $3x^2$ since $3xh+h^2$ gets close to $0$ and thus the derivative is $3x^2$.
Added: 
If someone is asking you what $\lim_{h\rightarrow0}3x^2+3xh+h^2$ is, they are asking you what will $3x^2+3xh+h^2$ come close to if you substitute for $h$ values very close to $0$ (but not $0$ itself). Take an example: Find $\lim_{x\rightarrow 2} \frac{(x+2)(x-2)}{x-2}$. Now try to put values for $x$ that are very close to $2$ and see what the function, $f(x)=\frac{(x+2)(x-2)}{x-2}$ gets close to. For $x=2.1$, $f(x)=4.1$. Similarly $f(2.01)=4.01$, $f(2.001)=4.001$, $f(2.00001)=4.00001$. Also for values below $2$, $f(1.9)=3.9$, $f(1.999)=3.999$, $f(1.999999)=3.999999$. It is evident that the function $f$ is approaching $4$, as you input values close to $2$. Note that the function is itself undefined at $x=2$, but that does not matter since the limit only cares about how the function behaves near $2$ but not at $2$. So, $\lim_{x\rightarrow 2} \frac{(x+2)(x-2)}{x-2}=4$
A: It is not 'f(x)=' but its 'as limit h approaches 0'. So, after simplification you put h=0 and thus, the other terms cancel out. The book shows the method of differentiating using first principle. 
A: In the early development of the calculus, mathematicians were interested in the effects of "small changes" in the domain ($x$ values) of a given function on the range of the given function (the values $f(x)$ which result)
A small change was called a differential, which we will write "$\text{d}x$". In some sense this can be thought of as the value $h$ you have above, and as $\text{d}x$ gets very small, or similarly $h$ gets very small, we understand it to "go to zero." This is the reason we ignore the terms with coefficients of orders of $h$ in your equation.
Here's an interesting example from history by Euler (adapted from Dunham 2008, "The Calculus Gallery," pp.53-54), which would not meet today's standard of proof, but might give some insight into what's going on:
Euler was interested in differentials of the sine function. He started with the power series of $\sin(x)$ and $\cos(x)$, thus
$$\sin x = x-x^3/3!+x^5/5!-\cdots$$ and $$\cos x = 1-x^2/2!+x^4/4!-\cdots.$$
He then substituted $\text{d}x$ into both (which he understood to be "infinitely small" - whatever that means!):
$$\sin(\text{d}x) = \text{d}x-(\text{d}x)^3/3!+\cdots$$ and $$\cos(\text{d}x)=1-(\text{d}x)^2/2!+\cdots.$$
The higher powers of $\text{d}x$ (such as $(\text{d}x)^2$ and $(\text{d}x)^3$ etc.) were considered insignificant (since they "are zero even more so than $\text{d}x$ or a constant" - again whatever that means!). Hence, we can write (*)
$$\sin(\text{d}x)=\text{d}x$$ and $$\cos(\text{d}x)=1.$$
Euler then considered the function $y=\sin(x)$, and using the idea of the differential considered:
$$y+\text{d}y = \sin(x+\text{d}x).$$
Using a trigonometric identity this gives
$$y+\text{d}y=\sin(x)\cos(\text{d}x)+\cos(x)\sin(\text{d}x),$$
which using (*) gives
$$y+\text{d}y = \sin(x)+\cos(x)\text{d}x.$$
Subtracting $y$ from both sides then gave
$$\text{d}y = \sin(x)+\cos(x)\text{d}x-y,$$
which simplifies to $$\text{d}y = \cos(x)\text{d}x.$$
In modern notation this tells us that $$\frac{\text{d}y}{\text{d}x} = \cos(x),$$
 i.e. the derivative of $y=\sin(x)$ is $\cos(x)$ !
