Tangent line to a graph and derivatives? My book states:

Assuming that $f(x)$ is differentiable at $a$, the tangent line to the graph $y = f(x)$ at the point $(a, f(a))$ is given by the equation.
  $$f(x) - f(a) = f'(a)(x - a)$$

Now let me ask my question.
$f'(a)$ is an instant change (when $\Delta x \rightarrow 0$) of the function $f(x)$. If such, I can't get why we multiplying it by the $(x - a)$? I assume you can take any $x$ and any $a$. Thus the distance between them might be quite big. Why the product of non-infinite-small distance and derivative is valid? What am I missing there? 
The only way I could agree on that, is when
$$f(x) - f(a) = f'(a)(x - a)$$
$$x \rightarrow a$$
Then the distance $(x - a)$ is infinitely small and thus derivative is applicable.
 A: The tangent line is a good approximation locally but it need not be a good approximation when $|x-a|$ is huge. 
To study the error term, If $f$ is of class $C^{2}$, then for each $x$, there exists a $c$ between $a$ and $x$ such that 
$$f(x)-p(x)=\frac{f^{(2)}(c)}{2}(x-a)^2$$
here $p(x)$ is the tangent line.
A: I am not very kind of the way your book writes things off. 
You have a function $f(x)$ for which a derivative exists in $a$, thus $f'(a)$ is defined.
Then if you want to have the equation of the tangent of the function at point $a$, you have the following:
$y(x)=f(a)+f'(a)\times (x-a)$
Why? Well, the tangent goes by the point $(a,f(a))$. On top of that, this is a line, for which the general equation is $y(x)=cx+d$.
Let's do $y(a)=c\times a +d=f(a)$. And the rate of the line is equal to the derivative in point $a$, thus $c=f'(a)$. You have then $d=f(a)-f'(a)\times a$.
Thus the original equation! $y(x)=f'(a)\times x+f(a)-f'(a)\times a$...
A: A bit more in detail.
1) Straight line:
Let the slope $m$, and a point $(a,b)$ given.
The equation: $y -b = m(x-a).$
a) Line passes through $(a,b)$ ;
b) Has slope $m$ . 
Check it.
Being given $f(x)$, $f'(x=a),$ is the 'slope ' $m$ of the tangent line passing through $(a,f(a)).$
If you accept the above statement you are done.
The more interesting question is, how do we get the 'slope ' of the tangent line to a curve $f(x).$
