does anyone know what he is talking about? definitely not me
Basically, it’s (y2 - y1)/(x2 - x1), right?
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Can you elaborate on this? Doesn’t it just approach zero the smaller and smaller you go?
Let’s do an example. What is the instantaneous rate of change at any point on the function, f(x) = x^2?
Well, let’s plug it in to our limit. Let’s just say Δx is h for easy purposes.
lim h → 0
(f(x + h) - f(x))/h
Since our function is f(x) = x^2, we turn the “function part” into this:
(x^2 + 2xh + h^2 - x^2)/h
Cancel x^2 out.
(2xh + h^2)/h
Divide h on the top and bottom.
(2x + h)/1
Now, since h approaches 0, we just plug in 0 normally.
2x.
So, the instantaneous rate of change or derivative on any point of the curve f(x) = x^2 is 2x
Check out the example.
Δx, or let’s say h becomes smaller and smaller, like 0.001. If we have (f(3 + 0.001) - f(3))/(3 + 3.001 - 3), we can get really close to finding the instantaneous rate of change of the curve at that point, because the difference between the two numbers is so small.
oh, ok i think i get it now. you see, i will not be in high school next year, and we have only worked mainly with linear functions, none with exponential factors
Well, if our function is f(x) = x^2, than f(x + h) is just (x + h)^2. I’m assuming you know that special case, so yeah.
I never thought I’d be explaining derivatives on a Gimkit forum, but here we are…
That’s why I deleted it
The notation of a derivative is usually dy/dx (the derivative of y with respect to x), d/dx [f(x)] (the derivative of f(x) with respect to x) or f’(x) (lagrange’s notation I think)
Here are some rules.
If f(x) = x^n, than f’(x) is nx^(n-1)
But as Delta x approaches zero, then it becomes (f(3)-f(3))/(3) → 0/3 → 0?
y = x^n
The inverse:
x = y^n
y = nroot(x)
Shouldn’t it be this?
Well you see, you can’t always evaluate limits like that. You have to look for more creative solutions, also it’s (f(3)-f(3))/(0), actually.
You can plug your “h” in normally, but if it gets you to 0/0 then you can use other methods, like the one I used. Remember, Δx never actually becomes 0.
That doesn’t really have to do with a derivative, and it would actually be n = logₓy
But if I nroot both sides, then I get:
nroot(x) = nroot(y^n)
nroot(x) = y
Is this wrong?
But that would mean this is wrong:
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Why?