we were studing the rate of the function
$\frac{f{x_1}-f{x_2}}{x_1-x_2}$
if it is positive so the fonction is growing if it is negative so the function is ascending .
in this moment our teacher tolds us that the derivative of a function have the same signification ( because we will study the lesson of derivation of a function the next year)
Then I was very curious I asked him to explain me it and here was the problem
He tolds us that what we were studing in physics is an application : the lesson of speed is a great exemple
the function : $V_{m}=\frac{d}{\delta t}$
Then he tolds us that
$V_{i}=\frac{M_{i-1} M_{i+1}}{t_{i-1}-t_{i+1}}$
Wich we can calculate the instantaneous speed is a derivative of the function
My question is : is there really a relation between the derivative of a function and the instantaneous speed ? How is that ?
Because I don't know even what is a derivative of a function ?
Why we are studing in phsics something we didn't study in mathematics?
Welcome for any suggestions edit for my spelling :) thank you
Let us define an function for your distance covered.
Name it $d=s(t)$ Here, we are assuming that distance covered is function of time.
Now what is instantaneous velocity?
Well let us suppose we have an interval of distance and an interval of time.
Now suppose that at $t=t_0$ Distance covered is $d_0=s(x_0)$
Now we add a small amount of time or assume that a very minute quantity of time $\Delta t$ has passed , now our total time is $t_0+\Delta t$
Meanwhile our distance has become? $d_{1}=s(t_0+\Delta t)$
Now we will find the average speed between the time interval we have mentioned which is given by?
$v_{avg}=\frac{d_1-d_0}{t_0+\Delta t -t_0} $
$v_{avg}=\frac{s(t_0+\Delta t)-s(t)}{\Delta t} $
Now I ask you what will happen when $\Delta t \to 0$ .
I will leave that for you but i can tell you how it's written.
$=\lim_{\Delta t \to 0}\frac{s(t_0+\Delta t)-s(t)}{\Delta t}= \left\{\frac{ds(t)}{dt}\right|_{t=t_0}$