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.
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I mean I think the other answers covered it but the thing that I always visualized was the tangent line. If you have a linear equation like F(x) = 5x.. then it doesn’t really make sense to draw a tangent line of course because it will be the same line.. but if you did that, the slope of the line would be 5. So the slope of the tangent line is your derivative (5).
Now, for a parabola, the tangent line is different at any given point, and gets steeper as the parabola trends upward. But the slope of that line will be your derivative. For the x^2 example, of course the derivative will always be 2x.. or put another way, the slope of the tangent line will always be 2x.
The reason I bring up the tangent line is because the tangent line only touches the parabola at a single point and is a different line with a different slope at every point. This drives the ‘point’ home that the derivative is about singular points along the line, not big jumps like from F(1) to F(2).
You get in your car, buckle up, turn it on, and put the pedal to the metal. The trip-o-meter started at 0 miles but quickly starts going up really fast, and soon everybody else is whizzing past you in a blur.
You may look at your speedometer and think it measures your speed, and it does. But it measures how fast you’re going even if your buddy takes a picture of you. While the picture doesn’t show you moving, per se, it does show you’re whizzing super fast cause your car is a blur. This picture of you, the glance at your speedometer is the derivative. It is an instant in time.
But if your buddy on the street wanted to measure how fast you were going they’d have to use a stopwatch and start it when you whizzed past the tree and then stop it when you flew by the lamp post. He’d then measure how many feet between those 2 points, then divide by the time. His measured speed didn’t match up with your speedometer between the tree and lamp post because you saw the indicator smoothly go from 0 to infinity and beyond. It only hit your buddy’s measured speed once. He would be wrong thinking you went this measured speed the whole time between the tree and lamp post.
His was an average over some range while your was instantaneous snapshot, just as your F(3)-F(2)=5 was different than F'(3)=6. Your speed probably started at 4 when you went past the tree and got to 6 when you past the lamp post, but this averages out to 5 from your buddy’s perspective. The instant you past the lamp post your speedometer showed 6.
What you’re describing in your third bullet-point is a line SECANT to the parabola that intersects it at exactly two points, at F(1) = 1 and F(2) = 4, and it has a slope = 3.
The derivative is the value of the line TANGENT to parabola at that point, that intersects is exactly once, THAT is what has has a slope = 4 in your example.
In your example, the tangent line is steeper than the secant line, that’s why it has the higher value. You can’t add and subtract values of a function and its derivative the way you were, that’s why the numbers aren’t lining up!
Ok so the derivative could be thought of finding the slope of a curve at a point, this is not exactly possible because the equation for slope is x2-x1/(y2-y1) if the 2 points are the same then you end up with 0/0 which is indetermanent, so what we can do is pick a point close to x=2 so let’s chose x = 2.1 for our second point. We end up with the points (2,4) (2.1,4.41) this gives us the slope of 4.1, not quite the same as the actual slope but close, so let’s choose a point closer to x=2, we will go with the points (2,4) (2.05,4.2025) we end up with the slope 4.05 which is more accurate but still not quite.
but we can see as x2 gets close to the value of x1 the slope between these two points is converging to a certain number, so let’s do the point x=2+(1/infinity) excuse the notation but this is roughly saying the value closest to x=2 but larger than 2 if this value were expressable the resulting slope would be 4, one way for us to compute this is by taking the limit as x2 goes to x1 but
What you are doing is apart of this but you are forgetting to move the other point along the curve.
This is how taking a derivative is done you are just attempting to do this process for all points of the curve that are differentiatable in the case of f(x) = x^2 the derivative or F'(x) = 2 *x. One way to check that this is the derivative is to create a tangent line using the function: t(p,x) = F'(p)*x+(f(p)-(F'(p)*p)) where p is the x value you want a tangent at this will give you a tangent line and will allow you to graph it, what you should notice is this line only touches the curve at the point of x=p and this will be true for all x where the function f(x) is differentiatable.
You might think of the derivative as an instantaneous average rate of change. That is, you find the average rate of change over a very small interval. You might think of how to estimate the speed of a car: calculate the average rate over a very small time period.
Think about it this way. Lets say you are in a rocket, and to measure your speed you start by measuring distance in meters from where you started. Coincidentally let’s say it’s a 10 second test, where your distance is defined by t². So at 1 second your 1 meter away, 2 your 4 meters, 3 your 9 meters away, etc. At the end of the 10 seconds your 100m away.
That’s great but your trying to measure your speed each second. So first you say well I went 100 meters in 10 seconds so 10 meters per second. ….But wait that’s not your top speed, your top speed is 20m/s.
For F(x) = x² => F'(x) = 2x you made a math error a little later… F'(1) = 2 not 1. Your subsequent reasoning about the rise from F(0) to F(1) being the same value got you more off track.
As others have stated, the result of the derivative is the current change in the function, not the average. The fact that we can compute the current change and not the average is the gift that calculus gave the world.
f(x) = x^2 is the first function in your examples that becomes curved – it’s no longer a straight line. So what changes from that? What changes is that when you go from x = 1 to x = 2 the rate of change “ramps up”. When you’re near x = 1 it’s still approx. 1, but at x = 1.1 it’s 1.21. In this way it ramps up all the way to 4 at x = 2.
So what can you learn from this? You should notice that there’s also lots of numbers, in fact infinite amount, between the numbers you pick. And they all have their own rates of change for such non-linear or curved functions as f(x) = x^2. I’d like you to correct your intuition about derivatives from them being a straight line between any two arbitrary picked numbers like you did, to instead it being a straight tangential line that merely touches the function at just a SINGLE point as far as you’re concerned.
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