I was thinking about why the numerical value of the derivative is helpful. Honestly, I find the numerical value a bit misleading with regards to the units of the derivative. I would love someone to help me clear up my confusion:
Let the position, in meters, of some moving object at time $t$, in seconds, be represented by $f(t) = t^2$ for $t>0$. Say $t_0$ seconds pass and now $f(t_0)=t_0^2$.
We know after some small, tiny change in seconds $dt$, the object will move forward by quantity $df(t_0)$ meters.
When evaluating the numerical quantity of $df(t_0)/dt$, the intuition feels obscured especially since $df(t_0)/dt$ is expressed in meters per second.
To compute the ratio of $df(t_0)/dt$, we are basically treating the numerator and denominator as legs of a right triangle (where $df(t_0)$ is the opposite side). We use the similar triangles property by scaling said triangle upwards so that the adjacent side (originally of length $dt$) now has length $1$. So our value of $df(t_0)/dt$ is the linear approximation of $f(t_0+1)$ (call it $l$).
Reading off the value of $df(t_0)/dt$ in meters per second makes one feel that one second later, the object will be at position $l$ meters when the object is really at position $f(t_0+1)$ meters. So even though the derivative is the instantaneous rate of change, the derivate's value reads as if we are moving the independent variable forward by one unit.
So why is the derivative useful in the real world if most changes in the independent variable occur in not-so-tiny steps instead of itty-bitty steps like $dx$? Scaling the derivative by a not-so-small quantity that is written in the same units as the independent variable's feels counter-intuitive.
I know in machine learning, a default learning rate is about $0.01$ and is consistent with this notion of "itty-bitty".
Any insight is appreciated, especially from the physics perspective.
EDIT: Not to completely ignore the responses to this question, but I think I understand what was tripping me up. If you are going to scale $df(t_0)/dt$ by some quantity $a$ where $a$ has no units, we are taking $a$ itty-bitty steps of size $dt$ and having the output of the function $f$ change by $a*df(t_0)$ (this make sense especially in regards to the directional derivative). Yet, if $a$ is written in the same units as the independent variable $t$, then those units will cancel out and we are left with the linear approximation of $f(t_0+a)$ which is written in the same units as the dependent variable. I take it that this latter situation is helpful if $f(t)$ is a linear function or $a$ is rather small, like an itty-bitty step.
