# 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 ?

• The derivative has an exact value and is roughly the change of the function if you barely move(with more precision closer you are to $a$). – ruler501 Mar 29 '14 at 4:44
• Could whoever downvoted the question care to explain why ? What about the question shows either lack of research effort, lack of clarity, lack of usefulness ? – curryage Apr 1 '14 at 10:21

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 ($\epsilon$-$\delta$)

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 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.

• I don't think $h$ is actually $0$. – curryage Mar 29 '14 at 5:13
• @curryage it is the limit as it goes to zero so it can be taken as being zero. You can also take it to be some infinitesimal value, but that is slightly more complicated. – ruler501 Mar 29 '14 at 5:14