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 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)
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