I know the definition of derivative is :

$$ \frac{df(x)}{dx} = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x} $$

But I'm confused about it . When we say $\Delta x\to 0$ , so $\Delta x \not= 0$ and the result is derivative is an approximation and it isn't precise . For example consider a car moves with $f(t) = t^2$ equation . So the velocity equation is $g(t) = 2t$ . When $t=5$ the velocity is $10 \ m/s$. According to derivative definition $v$ tends to $10$ but also we know it will get $10$ passes to $11$ , $12$ , $...$ .

If my question seems silly , I'm really sorry.

  • 1
    $\begingroup$ I don't think I understand what your doubt it,can please be more precise? $\endgroup$ – Fawad Mar 26 '17 at 12:29
  • $\begingroup$ Okay , In the definition of derivative we say $\Delta x\to 0$ so we never get precise value. $\endgroup$ – S.H.W Mar 26 '17 at 12:38
  • $\begingroup$ We get precise Vale. It would be better if you see it graphically. $\endgroup$ – Fawad Mar 26 '17 at 12:41
  • $\begingroup$ When we draw tangent line $\Delta x = 0$ but in the definition we say $\Delta x\to 0$. $\endgroup$ – S.H.W Mar 26 '17 at 12:46

Perhaps your confusion is that you're thinking of the symbol “lim” as denoting the process of taking the limit, whereas it actually denotes the result.

The value of $\Delta f/\Delta x$ is changing as $\Delta x$ is changing. So yes, that quotient doesn't have a definite value. But it approaches a definite number as $\Delta x \to 0$, namely the limit. The limit is just a fixed (precise) number $A$, sitting still at its position on the number line, waiting for the moving number $\Delta f/\Delta x$ to approach it.

Using an analogy, the limit is the destination of a traveller's journey, not the traveller or the journey itself. (And in this case the traveller may actually never reach the intended destination, just get arbitrarily close.)

It's this (precise) number $A$ which is called “the derivative”, and denoted by $f'(x)$ or $\frac{df}{dx}(x)$.


Some preliminary remarks. We have no ground to judge mathematical concepts by real-world phenomena; that you can test a physical theory w.r.t. real-world phenomena is because it asserts something about phenomena. If you are given an equilateral triangle, are the side lengths equal? To answer this question, you don't take a ruler to measure that triangle and report "oh, the side lengths are not equal based on my measurements (a real-world phenomenon, so to speak)." Why? Because an equilateral triangle has by definition got equal side lengths. It is in this sense I said that we have not ground to judge math concepts by real-world phenomena. In the same token, the definition of derivative is not a model for physical phenomena. It is applied to real-world phenomena; it is not tested by real-world phenomena. The only judge in math is logic.

You seem to have a predetermined sense of "precise". To say something is precise or not you must have a criterion there. Since apparently everyone more or less has implicitly or explicitly his own criterion for being precise, there is no way to discuss this concept properly.

The word "approximation" in math is used in a comparative sense. For example, if we want to calculate the area of a given triangle and if somehow the area of the circle inscribed to it is easier to calculate, then we use the area of the circle as an approximation to the triangle. Can we use the area of a triangle to approximate that of a circle? Yes, if to calculate the area of a circle is more difficult than to calculate that of the triangle inscribed to the circle, then we use the area of the triangle as an approximation to that of the circle. This is it. You saw that it is purpose-dependent to use the word "approximation". And, therefore, if there is a universal sense of "precise" adopted, there is still no way to say an approximation being precise or not until "what approximates what" is clear.

To understand the definition of derivative, you want to understand the concept of limit mathematically (instead of physically). If you understood the concept, then no worries. Take it with a simple sequence $(1/n)$. We have $1/n \to 0$ as $n \to \infty$. Here $n$ is never $\infty$ (there is no real number called $\infty$) and $1/n \neq 0$ for all $n$. Take a closer look at the definition you will know that it requires that for every $\varepsilon > 0$ there is some $N$ of such a kind that $n \geq N$ implies $|\frac{1}{n}-0| = \frac{1}{n} < \varepsilon$. In plain English, this is to require that you can make $1/n$ as close to $0$ as you please ($< \varepsilon$ that close) as long as $n$ is sufficiently large (greater than or equal to $N$). This is the rule of the game; no more and no less. As for derivative, likewise it requires that you can make the difference quotient as close to some number as you please as long as $\Delta x$ is sufficiently close to $0$.

  • $\begingroup$ I know definition of the limit , when we calculate a limit we care about neighbor of the $c$ not exactly $c$ . $\endgroup$ – S.H.W Mar 26 '17 at 13:53
  • $\begingroup$ Yes. I said so only because I suspect you know it without understanding it :). The concept of limit is interested in the behavior of a function near a point without necessarily being defined at that point! If you understand the concept, then no worries. $\endgroup$ – Megadeth Mar 26 '17 at 14:01

I should not add anything after Eric Clapton's answer.

Leu me try to illustrate what happens with your data using $$f(x)=x^2\qquad ,\qquad\frac{\Delta f(x)}{\Delta x}=\frac{f(x+\Delta x) - f(x)}{\Delta x}\qquad,\qquad x=5$$ anf let us use $h=10^{-k}$. Trying different values of $k$, we get $$\left( \begin{array}{cc} k & \frac{\Delta f(x)}{\Delta x} \\ 0 & 11 \\ 1 & \frac{101}{10} \\ 2 & \frac{1001}{100} \\ 3 & \frac{10001}{1000} \\ 4 & \frac{100001}{10000} \\ 5 & \frac{1000001}{100000} \\ 6 & \frac{10000001}{1000000} \\ 7 & \frac{100000001}{10000000} \\ 8 & \frac{1000000001}{100000000} \\ 9 & \frac{10000000001}{1000000000} \\ 10 & \frac{100000000001}{10000000000} \end{array} \right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.