Nader AbdlGhani said:
I have a textbook indeed, but it doesn't explain the concept clearly as it depends on how to get the derivative not what it is , for me getting the derivative of a function is pretty easy but personally I like to fully understand what I study so that's what got me here , feel free to show me any tutorial , thanks .
Nader, I sympathize with your ambition. Too many students learn how to calculate derivatives without understanding what they are doing. So I'm happy that you want to understand.
Now, at a quick search, I don't find any tutorials which contain what I would like, so instead, let me give you some exercises, based upon the following scenario:
You drop a rock from a high tower, and it falls freely. After ##t## seconds it has fallen ##y(t)=4.9 t^2## metres.
1. Calculate the distance the rock has fallen after ##3## seconds. that is: ##y(3)##.
2. Calculate the distance the rock has fallen after ##4## seconds, that is: ##y(4)##.
3. Continue to calculate ##y(3.1)##, ##y(3.01)##, and ##y(3.001)##. Write down the values in a table.
4. Calculate the distance the rock has falls between ##t=3## and ##t=4##, that is ##y(4)-y(3)##.
5. Continue to calculate ##y(3.1)-y(3)##, that is, the distance the rock falls between ##t=3## and ##t=3.1##.
6. Similarly, calculate ##y(3.01)-y(3)## and ##y(3.001)-y(3)##. Write down all this in a new column in your table.
7. Calculate the average speed of the rock between ##t=3## and ##t=4##, that is ##\frac{y(4)-y(3)}{4-3}##.
8. Continue to calculate the average speed of the rock between ##t=3## and ##t=3.1##, that is ##\frac{y(3.1)-y(3)}{3.1-3}##.
9. Similarly, calculate the average speed of the rock between ##t=3## and ##t=3.01##, and between ##t=3## and ##t=3.001##. Write down the average speeds in yet a new column in your table.
10. Now if you look at your table, what would you say happens with the average speed over the time interval ##[3,t]## when it becomes smaller and smaller, that is, when ##t## approaches ##3## (from the right)? You may test your guess by making similar calculations for ##t=3.0001##, ##t=3.00001##, etc.
11. You could make a similar table to see what happends if ##t## approaches ##3## from the left instead, that is, using ##t=2##, ##t=2.9##, ##t=2.99##, etc. instead. Does the average speed approach the same value in this case too?
12. You should by now have concluded that the average speed of the rock over a time interval which has ##3## as its left or right endpoint approaches a particular value when the interval gets smaller and smaller, that is, when the length
approaches ##0## (but never
becomes ##0##, because then the denominator becomes ##0##, so we cannot caluculate the average speed in that case). So this value is the
limit of the average speed at ##t=3## as the time interval length approaches ##0##.
13. This limit of the average speed at ##t=3## is precisely what we mean with the
instantaneous speed of the rock at ##t=3##, that is, the instantaneous speed is
defined as this limit. If we could glue a working speedometer to the rock, it would show this very value at ##t=3##. Let us denote this instataneous speed as ##v(3)##.
14. You could make similar tables to estimate the intantaneous speed at other times, e.g. ##t=1##, ##t=2##, ##t=4## etc. to find ##v(1)##, ##v(2)## etc.
15. If you have learned how to calculate limits, you could even calculate the instantaneous speed at a general time: ##v(t)##, getting results with certainty instead of just guesses based upon tables.
16. Now, apparently, you have learned how to calculate derivatives of simple functions, so you can calulate the derivative ##y'(t)## of ##y(t)=4.9t^2##.
17. Compare the derivative of ##y(t))## at ## t=3##, i.e. ##y'(3)##, with ##v(3)##. Conclusion? If you could calculate ##v(t)## for a general ##t## at step 15, you can affirm your conclusion.
18. Some notation: at time ##3##: ##\Delta t = t-3##. ##\Delta y= y(t)-y(3)##. Alternative notation: ##h=t-3##, so ##\Delta t=h## and ##\Delta y= y(3+h)-y(3)##. The derivative can be written as ##v=\frac {dy}{dt}=\lim_{\Delta t\to 0}\frac{\Delta y}{\Delta t}## (at time ##3## and at any times).
19. ##v(t)=y'(t)## is itself a function of ##t## which also has a derivative, which also can be calculated by simple rules for differentatiation (in this case) or by calculating limits, and it can be estimated by tables like these above. By definition, the derivative of ##y'(t)## is ##y''(t)##. It is called the
second derivative of ##y(t)##. It is, by definition, the
acceleration of the rock at ##t##. We can continue and get the third derivative, the fourth, etc.
20. In this example, we had time (##t##) as the independent variable. Speed is the derivative (rate of change) of the distance with respect to time, and acceleration is the derivative (rate of change) of speed with respect to time. This is perhaps the most illustrative cases, but it needs not be so. Often one has ##x## as the independent variable, and we can in principle use any letter. If we blow up a baloon, we could take the derivative (rate of change) of the volume with respect to the radius. We can take the derivative (rate of change) of the tax with respect to the income, etc.