Can one interpret the derivative of $f(x)=x^2$ as a way to predict a value under which $f(x)$ cannot fall at a given point? and $f'$as a minimum rate? Note : my questions is not about the application of derivatives to determine the max / min values of a function ; also, t is not about linear approximation ( though it is related to this topic); what I am interested in is a possible alternative definition of the derivatve of a function

If P and Q are two points on the graph of $f(x)=x^2$, with $P=( x_0, f(x_0))$ and $Q = ( x_0 + \Delta x, f(x_0+\Delta x))$ the change in $y$ related to the change in $x$  is :
$R = \frac {f(x_0+\Delta x) - f(x_0)}{\Delta x}= 2x+\Delta x$.
If I am asked what is the value of $f(3,1)$ knowing that the value of $f(3)=9 $ I can use the ratio $R$ and write :
$f(3,1) = f(3)+ (\Delta x.R) = f(3) + (\Delta x .(2x+\Delta x))= f(3)+(\frac{1}{10}.((2\times 3)+\frac{1}{10}))= 9,61$ .
Now, if I were asked, not what is $f(3,1)$ but the greatest value under which $f(3,1)$ cannot fall I could take the limit of $R$ as $\Delta x$ goes to $0$  and say that $f(3,1)$ cannot fall under the value :
$f(3)+(\Delta x . 2x)=9,6$.
In other words, even  if I can't manage to find precisely $f(3,1)$, I can predict that it must be greater than  $9,6$.
So the derivative at $x_0=3$ gives me the minimum value of the rate $R= \frac{\Delta y}{\Delta x}$ at $3$, that is $6$ .
My question is : (1) is it correct to view the derivative as a minimum ( or maximum) not of function f itself but of the rate $R$ ( as defined above)? (2) how to make this idea of minimum / maximum rigorous? (3) is the notion of supremum needed here?
In bref, is there an alternative definition of the dervative that embeds these notions of maximum / minimum/ supremum?
 A: Your basic premise is wrong, I'm afraid. The only reason it works here is that the derivative $f'(x)=2x$ is increasing as $x$ increases. If you try it with a function that has a decreasing derivative, it doesn't work. For instance, if $$f(x)=-x^2+12x-18$$
then we have $f(3)=9$ and $f'(3)=6$, just as in your example; but $f(3.1)=9.59$, which is less than $9.6$ because the derivative $f'(x)=-2x+12$ is decreasing as $x$ increases.
A: [Comment promoted to answer at request of OP]
The graph of $y=x^2$ lies above the line tangent to the graph at $x=3$ (in fact, at every value of $x$), which is why your lower bound for the growth of the function serves as the derivative of the function. If the graph lies above the tangent line (but the function is still increasing), you'll be looking at an upper bound for the growth as the derivative. And you have to modify everything if the function is decreasing. And another ballgame if the graph lies above the tangent line in one direction and below it in the other. Essentially, you have to take the second derivative into account.
