Is this explanation of derivatives correct?

Question

Say there is a function $f(x)=x^2$

Now take any point $X$ on the line. This will correspond to coordinates $(x,y)$. Now take a small change $dx$ to the left and add it to $x$.

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This would mean that the co-ordinates of the new point would move to such that $x_1=x+ dx$ and similarly for $y_1=y + dy$

Difference in these two $y$ or $dy$ equals $$ \begin{aligned} (x+ dx)^2 - x^2 &= x^2-x^2+dx^2+2x\cdot dx \\ &= dx^2+2x\cdot dx \end{aligned} $$

Now as we make $dx$ smaller and smaller $dy$ also gets smaller. We could do this and stop just before the point $X$. Now change in $x$ would be only the point $x_1$ itself I.e. $dx = \text{Point } x_1$.

Since this is a $0$ dimensional quantity, we deduce the following:

$dx^2$ will become 0 as a point multiplied by itself is the point itself(as it is in 2 dimension, I am not saying that the point itself has no value).

Now $dy/dx=2x$ or for every unit time we experience a change of $2x$.

Questions:

1. Is my use of $dx$ as a point right?

2. Is my postulate that the distance between two consecutive points is the smallest value there is on a line correct? And if so is it what mathematicians mean by instantaneous rate of change?

3. Following 2. does this sidestep the paradox of how there exists instantaneous change by describing this change with respect to another point I.e. the point immediately next to it?

4. So Finally is all a derivative is, is a formula derived from differences between two $y$ values by using the fact that they all have a particular way of increasing (ex. for function of area of circle they increase by its perimeter no matter what $x$ is)?

Answer 1

Maybe this explanation can provide an intuition for the idea of " instantaneous velocity". I do not make any attenpt at rigor in this answer , for rigor in Analysis exceeds my ability.

Suppose that an object moves in a straight line under the action of a force, meaning that the object accelerates, and that its velocity is changing. The question " what is the instantaneous velocity of this object at time $t$? " can be rephrased as : "at what constant velocity would the object continue to move in case the force instantaneously stopped to act on the body? " . The constant velocity in this counterfactual scenario is what is meant by " instantaneous velocity".

Suppose that, under the action of the force, the function describing the movement of the object is $D(t)= t^2$.

In case the force suddenly stopped to act on the body at a given time $t_0$, the line representing the distance with respect to time would be, after time $t_0$ , the tangent to the graph at the point $P= (t_0, t{_0}^2)$.

And the derivative at $t = t_0$ is the slope of this tangent line.

NOTE : a mistake in the equation of the tangent due to a confusion regarding the variable $t$ ( and not $x$) ; so the correct equation is $y = D'(t_0) ( t-t_0) + D(t_0) $ .

As to $dt$ it is not supposed to denote a point but an " infinitesimal quantity", a very small change/ difference in $t$. ( Modern Calculus , since the precise definition of a limit, does no longer make use of this idea of "infnitesimal quantity").

As you noted $\Delta y$ / $\Delta t = 2t + \Delta t$.

When $\Delta t$ goes to $0$, without ever reaching this value, the quotient $\frac {\Delta y} {\Delta t}$ goes to $\frac {dy} {dt} = 2t$ , that is " tends to" $2t$.

The time interval is never actually equal to $0$, but, nevertheless it can actually tend to $0$ and in that case , $\frac {\Delta y} {\Delta t}$ actually tends to $2t$ without being ever equal to it ( unless the force stopped acting on the object, as in the counterfactual scenario I presented above).