Is the derivative an approximate value? From the article on derivative in Wikipedia:

The derivative of y with respect to x at a is, geometrically, the slope of the tangent line to the graph of f at (a, f(a)). The slope of the tangent line is very close to the slope of the line through (a, f(a)) and a nearby point on the graph, for example (a + h, f(a + h)). These lines are called secant lines. A value of h close to zero gives a good approximation to the slope of the tangent line, and smaller values (in absolute value) of h will, in general, give better approximations. 

Does this mean that derivatives of functions are just approximations ? (In the sense that the value of $f(x)$ at $a$ is exact -- $f(a)$ -- whereas the derivative $f'(a)$ is not so ?)
EDIT: Based on some of the comments, is it the case then, that the slope of the line passing through $(a,f(a)$ is the derivative and the approximations defined by the secant line slopes approach this real derivative in the limit ?
 A: No, the derivative $f'(a)$ is the exact slope of the line tangent to $(a, f(a))$.  Note that by definition
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
What the article is saying is that for very small $|h| > 0$, the slope of the secant line passing through $(a, f(a))$ and $(a + h, f(a + h))$ is a good approximation of $f'(a)$.  
A: The secant lines are the approximations. But if you take the limit of the approximations, you get the tangent line, which is exactly the true slope.
As a real world analogy, think of a moving car. To tell how fast it is going at time $t_0$, you measure out some small time $\Delta t$, and check how far you went, $\Delta x$. But your speed may have changed in the middle of your measurement! So you didn't get the exact speed at $t_0$. But if you make $\Delta t$ smaller, you'll get closer to the true value, right? So we take the limit as $\Delta t$ goes to $0$ to get the exact speed.
The limit of approximations is exact, in the same way that the limit of $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \ldots$ is exactly $1$. None of the actual terms are $1$, but the limit is exactly $1$, by definition of the limit ($\epsilon$-$\delta$)
A: 
A value of h close to zero gives a good approximation

In the case of a derivative, h is not merely “close to zero”, but an actual $0$. So the error is $0$ as well.
