When taking subsequent derivatives, why are units squared? $F(t)$ is a rate function measuring widgets per minute, its average rate of change goes to the next level derivative ... hence widgets per min$^2$
In related rates, the area is (unit$^2$) and volume is (unit$^3$).   I did not think of those as derivatives, but just 2D vs 3D calculations.
Let me think about the standard Distance/Velocity/Acc. relationship.


*

*$S(t)$ = distance traveled (in miles), $t$ = hours

*$S(1)$ = 50 miles.

*$S(3)$ = 150 miles.


So, Avg. Rate of Change over the interval $[1,3]$ is
$$ \frac{100 \text{ miles}}{2 \text{ hours}} = 50 \frac{\text{miles}}{\text{hour}}.$$
In this case, I have not squared the hours units b/c I started with an amount function (not a rate function)
Units make intuitive sense:  "50 miles for every hour."
But, let's say we consider the acceleration function, the derivative of the rate function.


*

*$v(1) = 40$

*$v(3) = 60$


The Avg. Rate of Change of velocity from [1,3] is 20 miles / 2 hours 
Is this why the Avg. Rate of Change of velocity should be written as 10 miles/hour^2?
In a nutshell,


*

*$S = \text{miles}$

*$S' = V = \text{miles/hour}$

*$S'' = V' = A = \text{miles/hour}^2$

*$S''' = V'' = A' = \text{miles/hour}^3$ ?


I do see the connection b/w the hour units and derivative level of S.
But, is it better to think about this as fractions?
$$S'' = V' = A = \text{miles/hour}^2$$
Does some dimensional analysis thing ever happen?
$$\frac{\text{miles}}{\text{hour}} \times \frac{1}{\text{hour}}
 = \frac{\text{miles}}{\text{hour}^2}$$
Here is where I arrived:  If you have a rate function, for example "miles/hour".   Then you take the rate of change of that?  "mile per hour, per hour"  or $$\frac{\frac{miles}{hour}}{hour} = miles/hour^2$$.  This also plays out in the average rate of change calculation:  $$\frac{f(b) - f(a)}{(b-a)} = \frac{(miles/hour - miles/hour)}{(hour - hour )} = \frac{(miles/hour)}{hour} = \frac{miles}{hour^2}$$
 A: The derivative of a function of one variable measures instantaneous rate of change, which has the same units as average rate of change:
$$
\frac{dy}{dt} = \lim_{\Delta t \to 0} \frac{\Delta y}{\Delta t}.
$$
Hence, the units of $\frac{dy}{dt}$ are simply the units of $y$ divided by the units of $t$.
A: Well it's really the only way to make sense of it. Velocity (I will use m/s) and acceleration ($m/s^2$). The reason the square goes in acceleration is to literally say it's gaining a velocity x m/s every second. Or, $x*(m/s)/s=x*m/s^2$.
All the next level does is add a "per unit" aspect.
A: The notation sort of suggests the units of the parameter should be exponentiated (and divided):
$$
f^{(n)}(x) = \frac{d^n}{dx^n}f(x)
$$
$dx^n$ will have the units of the $x$ to the $n^\text{th}$ power.  But $d^n$ is unit-less so, the final units would be the units of $f$ divided by the units of $x$ to the $n^\text{th}$ power.  A better way would be to understand what the derivatives are:
$$
\frac{df}{dx} \tilde{} \frac{\Delta f}{\Delta x} \\
\frac{d^2f}{dx^2} = \frac{d}{dx}\frac{df}{dx} \tilde{} \frac{\Delta \left(\frac{\Delta f}{\Delta x}\right)}{\Delta x} \tilde{} \frac{\Delta(\Delta f)}{\Delta(\Delta x)\Delta x}
$$
You can keep going and see that each time, you'll get more and more $\Delta x$ terms on the bottom, but the top just remains $\Delta(\Delta(\Delta ... (f)))$.
A: Here is where I arrived:  If you have a rate function, for example "miles/hour".   Then you take the rate of change of that?  "mile per hour, per hour"  or $$\frac{\frac{miles}{hour}}{hour} = miles/hour^2$$.  This also plays out in the average rate of change calculation:  $$\frac{f(b) - f(a)}{(b-a)} = \frac{(miles/hour - miles/hour)}{(hour - hour )} = \frac{(miles/hour)}{hour} = \frac{miles}{hour^2}$$
