Rate of a rate question (Calculus) The amount of sugar in a persons blood $x$ hours after drinking a soda is $f(x) = 80e^{-0.5x}$ mg. At what rate is the rate of the sugar in a persons bloodstream changing after $5$ hours?
I understand that the rate of the rate is the second derivative. Here $f''(x) = 20e^{-0.5t}$. But when I put $5$ in, I get about $1.64$. Mathematically I get it but, I don't understand what $1.64$ means application wise. Naturally the amount of sugar decreases over time so why is this not a negative number?
 A: First, lets look at the first derivative.  After $5$ hours, the rate at which the amount of sugar in the blood is changing is $f'(5)$, as measured in $\frac{\textrm{mg}}{\textrm{hr}}$ (this will be some negative number, indicating that the amount of sugar is decreasing  or that we are losing sugar).  Easy enough to understand.
Now, the second derivative measures how this rate is changing.  As you mentioned, the value of $f''(5)$ is $1.64\frac{\textrm{mg}}{\textrm{hr^2}}$  what this means is that, even though the amount of sugar is decreasing, the rate at which we are losing sugar is decreasing. Since the rate $f'(x)$ is negative we are losing sugar, but the rate of the rate $f''(x)$ is positive so the rate at which we are losing sugar $f'(x)$ is getting closer to $0$.  Hence we are losing less and less sugar as time passes
Disclaimer This is all purely theoretical.  According to these results, the instant you drink a soda, you have sugar from it forever, which is not true (at least I think not, but I'm no biologist).
A: Every time you differentiate with respect to time, append "per unit time".  For example, a unit of acceleration (the rate of the rate of change in position) is "meters per second per second".
Thus, your $f''(x)$ gives the amount of sugar "per hour per hour".  E.g., $f''(5)=1.64$ means that, starting from $x=5$ and over the next very short time duration $dx$, the rate of breaking down the sugar changes from $f'(5)$ (units of sugar per hour) to $f'(5 + dx) = f'(5) + dx * 1.64$ (units of sugar per hour).
A: As Andre pointed out, you are correct that the rate is negative, but the rate of the rate is positive here.  Note the function below in red, its derivative in green, and the second derivative in orange:

Recall that the derivative is very much related to slope, and slope is rise/run which gives you units in this case of $$\frac{mg}{h}$$  If you now find the slope of that slope then you have units of 
$$
\frac{\frac{mg}{h}}{h} \text{ or } \frac{mg}{h^2} 
$$
This last unit is not the rate of change of sugar, but the rate of change of rate of change of sugar.  That can be confusing, but consider that the first derivative is becoming "less steep" as time goes on.  
