Determining which function grows the most over time I have five logarithmic functions, 1 for each country I'm investigating. x = an integer value for years, y = share of 1-year-olds vaccinated against rotavirus.
So if x = 2, y would tell you how many 1-year-olds have been vaccinated against rotavirus (%) in the country's second year of progress.





Time Interval of x (Years)
a
b
y = a + bln(x)




Morocco
1-9
32.4
33.8
y = 32.4 + 33.8ln(x)


Sudan
2-9
22.3
34.7
y = 22.3 + 34.7ln(x)


Rwanda
3-9
41.8
27.9
y = 41.8 + 27.9ln(x)


Ghana
3-9
36.6
28.3
y = 36.6 + 28.3ln(x)


Botswana
3-9
75.4
6.18
y = 75.4 + 6.18ln(x)




This is for a school project, and I'm interested in determining which function grows the most over the course of its time interval -> so which country makes the most progress in vaccinations for the years of progress recorded.
I'm totally new to calculus but I want to include a calculus concept to determine this. I was looking into derivatives for calculating rates of change but I'm not sure if that's the simplest and most effective approach for what I want to do - would appreciate a suggested calculus approach potentially other than this that isn't too difficult to learn as a beginner!
 A: Given a function of the form $f(x) = a + b \ln(x) $, we can break it up into a couple of smaller functions.
First, the derivative (the function that describes the instantaneous rate of change) of the sum of two functions is the sum of their derivatives.
So, we can break up the function into the sum of two smaller functions. Let $f_1(x) = a$ and let $f_2(x) = b \ln(x) $.
Then, we have
$$ f(x) = f_1(x) + f_2(x) $$
$$ \Rightarrow f'(x) = f_1'(x) + f_2'(x) .$$
Here, the prime notation denotes the derivative of the function, so $f'(x)$ is the derivative of $f(x)$.
Now, $f_1(x) = a$ is a constant function - graphically, it's a horizontal line through $y = a$.
The rate of change of that function between any two points is always, since $y_2 - y_1 = 0$ for any two points on the graph. For that reason, the instantaneous rate of change is also 0 everywhere.
That gives us
$$ f'(x) = f_1'(x) + f_2'(x) = 0 + f_2'(x) = f_2'(x) . $$
Now, we need the derivative of $f_2(x)$. The derivative of the natural log function is $\frac{1}{x}$. This is in fact the way some mathematicians choose to define the natural log (and, by extension, the number $e$).
We can also use the fact that scaling a function by a multiple also scales its derivative in the same way. Given a constant $c$ and a function $g$, we have
$$ \frac{d}{dx}cg(x) = c \frac{d}{dx}g(x) . $$
Here, $ \frac{d}{dx} g(x) $ is another way to denote the derivative of $g(x)$ - it's equivalent to $g'(x)$.
This gives us:
$$ f_2'(x) = \frac{d}{dx} b \ln(x) $$
$$ = b \frac{d}{dx} \ln(x) $$
$$ = \frac{b}{x} . $$
So, putting this back into our equation above, we have
$$ f'(x) = \frac{b}{x} . $$
Notice, this is independent of $a$, since all $a$ does in our function is shift the curve up or down - it doesn't affect the actual rate of change of the function. That's why the derivative of the constant term was 0.
Now, applying this to the problem at hand, we want to know the instantaneous rate of change in year 9 (Matthew's advice for how to incorporate calculus). So, all we have to do is plug the value $ x = 9 $ into our formula for $f'(x)$.
For example, in the case of Morocco, the instantaneous rate of change in year 9 is:
$$ f'(9) = \frac{33.8}{9} . $$
$$ \approx 3.76 . $$
You can interpret that to mean that in year 9, the rate at which $y$ is increasing is approximately 3.76 percentage points per year. So, by year 10, you could expect that the proportion of one-year olds vaccinated in Morocco would increase by 3.76 percentage points.
