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
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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