does anyone know what he is talking about? definitely not me
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replying…
oh no this is taking a while don’t be zypheir again
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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?
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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
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Check out the example.
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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.
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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
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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.
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I never thought I’d be explaining derivatives on a Gimkit forum, but here we are…
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That’s why I deleted it 
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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)
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Here are some rules.
If f(x) = x^n, than f’(x) is nx^(n-1)
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But as Delta x approaches zero, then it becomes (f(3)-f(3))/(3) → 0/3 → 0?
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y = x^n
The inverse:
x = y^n
y = nroot(x)
Shouldn’t it be this?
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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.
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That doesn’t really have to do with a derivative, and it would actually be n = logₓy
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But if I nroot both sides, then I get:
nroot(x) = nroot(y^n)
nroot(x) = y
Is this wrong?
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But that would mean this is wrong:
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Why?
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