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
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.
Latest Answers