Ok so derivative are an expression of the rate of change of a function. Cool I get that.
– F(x) = 5 : the product of this function is always 5 there is no increase or decrease so there is no change no matter what X is and it makes sense that the derivative would equal 0.
– F(x) = 5x : it is obvious that each time x increases by 1 the product of this function increases by 5. I get it.
– F(x) = x² => F'(x) = 2x : starting from here the numbers stop matching and make me feel like I am missing something. F'(1) = 1. This makes perfect sense. F(x) did in fact increase by 1 when going from F(0) to F(1). Then I try F'(2) = 2×2 = 4. Huh ? But F(x) only increased by 3 between F(1) and F(2) ? Maybe I am looking at the rate of change as compared to F(0) ? after all there is an increase of 4 between F(0) and F(2). Let’s check with 3 then. F'(3) = 6. Wtf ?!
I don’t get it what does it mean when F'(2) = 4 ? When X = 2 then …? and what does it tell me about the original function. Thanks and hope my english isn’t too awfull.
In: 40
The derivative function helps us see when the original function is increasing (derivative positive) or decreasing (derivative negative). You can graph this to see.
Rate of change for this is also instanaeous. You averaged the change, that doesn’t tell you how much the function is currently changing.
The thing you are missing is that derivatives measure the instantaneous rate of change, not an average rate of change. By average rate of change, I mean (F(1)-F(0))/1 is 1; the average rate of change from 0 to 1 is 1.
You are correct that if f(x)=x^2, then f(2)-f(1)=3 not 4. So the average rate of change from 1 to 2 is 3. But what you have missed is that f(3)-f(2)=5. The average rate of change on the other side of 2 is bigger! That is, the rate of change of this function increases as x increases.
Now, to make precise the notion of “instantaneous” rate of change goes a bit beyond ELI5, but I hope it makes sense that if x=1 to x=2 has an average rate of change of 3 and x=2 to x=3 has an average rate of change of 5, the instantaneous rate of change at x=2 should be in between those numbers!
EDIT: removed exclamation point; no factorials!
If you make the increment smaller, you’ll see it approach the derivative, so instead of 1-2-3, try 1.9-2.0-2.1. then try 1.999-2.000-2.001.
The derivative is the instantaneous change, not the average change between two points.
When x=2 the function is changing at a rate of 4. But as soon as it goes above 2 the rate increases.
Strictly speaking you can’t use the rate of change over anything other than an infetissimally small change, i.e. instantaneous. As you’ve seen if you try to use it over larger distances it introduces errors.
You could also visualise it as the rate of change is changing. If you differentiate it again you’ll get 2, which is obviously not 0.
You are looking at the slope (rate of change) of the function at intervals of 1, but derivatives are all about the rate of change over an *infinitesmally tiny* interval.
Repeat your experiment, calculating the change in f divided by change in x. with the following values:
x goes from 0 to 0.001
x goes from 1 to 1.001
x goes from 2 to 2.001
x goes from 2 to 2.0000001
And you’ll see what’s going on.
The thing is that the derivative of a function at one point as sense only if you observe the variation close to this point.
The exemple you give of F(x) = x² with F'(2) give you information on the evolution of F(x) very very close to 2. So going from 1 to 2, or 2 to 3 is a very big step, too big to still have a meaning closely linked with the derivative in 2.
I can advice you to see a nice video on youtube made by 3Blue1Brown on derivative “Le paradoxe de la dérivée | Chapitre 2, Au coeur de l’analyse”. I know the title is in french, but the video is in english and it’s explain very nicely what a derivative mean.
The derivative tells you the instantaneous rate of change at the point where it is evaluated. This isn’t relevant when the rate of change is 0 (as when F(x) = 5) or when it is constant (as when F(x) = 5x). However, the rate of change is *increasing* when F(x)=x^2. You can see this if you plot the function. It starts out quite flat but gets steeper for higher values of x.
So F'(2) is telling you the rate of change when x=2, but the change between 1 and 2 depends on all the values of x between 1 and 2. The derivative of the function at these values will be less than 4, and it turns out their total contribution is to increase the value of the function by 3. You can see this by taking the simple average of these values:
[F'(1) + F'(2)]/2 = [2+4]/2 = 3
(Note that this strategy of using the endpoints to take an average only works with a linear derivative like this. To get more general, you’d have to start working with integrals)
> F(x) only increased by 3 between F(1) and F(2)
– F(x) increases by 3 between F(1) and F(2). The slope is 3/1 = 3.
– F(x) increases by 0.39 between F(1.9) and F(2). The slope is 0.39/0.1 = 3.9.
– F(x) increases by 0.0399 between F(1.99) and F(2). The slope is 0.0399 / 0.01 = 3.99
– F(x) increases by 0.003999 between F(1.999) and F(2). The slope is 0.003999 / 0.001 = 3.999.
As you bring the two points close together, the slope approaches 4.
How much did F(x) increase between 1 and 2? 3.
How much between 1.9 and 2? 0.39. But we should adjust for the smaller change in x, by dividing by 2 – 1.9 (= 0.1), to get 3.9.
How much between 1.99 and 2? (Make the right adjustment.) 3.99
How much for between 1.999 and 2? 3.999.
Notice a pattern here? What happens as we take smaller and smaller intervals to analyze? The answer narrows down on a certain value. That’s what the derivative of a function is.
Let’s take a look at F(x) = x^(2)
Pick a point on that curve. Let’s say (0,0). Now let’s pick a “distance” of 1 to the right of our x-value. Our x-value was 0, so now we’ll look at x = 1 which is also 1. So our second point is (1,1).
The slope of the straight line between these two points is 1. But you’ll notice that this straight line isn’t our curve. So the slope of that line can’t be the slope of our curve. To make it match our curve better, let’s pick a shorter distance. Let’s say 0.5. This translates to a point of (0.5, 0.25) and a slope of 0.5. But, again, this straight line doesn’t exactly match our curved line.
What we want is to make our distances smaller and smaller and see if the slope converges on some value. Then we will call the slope of our curve at that exact single point whatever that convergence is. If our distance is labeled “h” what we want to know is:
The limit of f(x+h) – f(x) / h as h goes to 0.
Obviously we can’t just plug in h = 0 but we can do some fancy math:
(x + h)^(2) – x^(2) / h
x^(2) + 2xh + h^(2) – x^(2) / h
2xh + h^(2) / h
2x + h
Now that we no longer have an h in the denominator we can plug in 0 for h and get 2x.
So, as h approaches 2 the slope will approach 2x. So, for our purposes, we simply say the slope at that exact point equals that limit of 2x. So, at x = 0, the slope of the curve (and also the slope of a straight line tangent to the curve at that point) is 2*0 = 0
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