# My (Weird) Way to Prove Differentiation (of y=ax^2+bx+c)

# Foreword

So I independently discovered this a few months ago.

(Also haven’t used hexo for way too long that I almost forgot how to write one of this)

(I am not going to try to get formula display on this site)

# The Thing

So, we have y=ax^2+bx+c, a simple general quadratic function.

To prove its derivative most people would just go for the first principle. Well, this would be kinda like it, but a bit different.

Let’s start by first factorise the x out of the ax^2+bx

y=(ax+b)x+c

Then, what does this looks like?

y=mx+c of course.

But you would probably argue m is not a constant if we simply equate these 2, that’s ture.

Still, let’s try to figure out what m actually is, although it’s not the gradient of the graph at x=x, it still must be a gradient (this was my gut feeling).

Did you figure it out? m is actually the gradient of a straight line passing the y-intercept and the point at (x,y).

Cool and now we do the first principle thing, reduce the distance between y-intercept and (x,y).

So, basically, approach x to 0.

And the m now becomes a*0+b or simply b, and this means the gradient of a quadratic at x=0 is b.

Now the problem is, we also want m of other points. It’s actually quite a simple solution, to just translate the graph horizontally, to make that other point the y-intercept.

And from now on I’m just gonna straight copy from my notebook:

i.e. for point (p,ap^2+bp+c)

to make the point the y-intercept:

x => (x+p)

y=(ax+b)x+c => (a(x+p)+b)*(x+p)+c

Which can again be put into the form y=mx+c as follows:

y=(a(x+p)+b)(x+p)+c=(ax+ap+b)(x+p)+c=(ax+ap+b)x+apx+ap^2+bp+c=(ax+2ap+b)x+(ap^2+bp+c)

Therefore as x becomes 0, the gradient of the point (p, ap^2+bp+c) on the original curve becomes 2ap+b.

So dy/dx=2ax+b