Derivative is just speed of change? In school we've been told that derivative of $x^2$ is $2x$.
Also I've read that derivative is simply a speed of value change.
So if
$$f(x)=x^2$$
then, using simple explanation, derivative of that function would be
$$f'(x)=f(x+1)-f(x).$$
Now if we take derivative when $x=2$ we will get
$$f'(2)=f(2+1)-f(2)=3^2-2^2=9-4=5.$$
But if we will take conversion rule (from school) which says that $[x^2]'$ is $2x$ then
$$f'(x)=2x$$
and if we will put the same point here we will get
$$f'(2)=2*2=4$$
so first result gives me $5$ and second result gives me $4$. And this problem seems to be appearing for every number. Number calculated by simplified interpretation is always bigger by $1$.
And my only guess is that simplified explanation missing something. Or, maybe, I made a mistake somewhere.
Can you, please, help me figure it out?
Update:
I spend 2 days trying to figure this out! Thanks to all of you, guys!!! Now i got it!))))


then, using simple explanation, derivative of that function would be
$$f'(x)=f(x+1)-f(x).$$

Actually, no. What you describe would not be the speed of the change, it would simply be the change. The speed of the change would have to be "how much change happened" divided by "how long the change took place". So, the speed of change on an interval of length $1$ would be $$\frac{f(x+1) - f(x)}{1},$$ but the speed of change on an interval of length $0.5$ would be $$\frac{f(x+0.5)-f(x)}{0.5}$$ and in general, the speed of change on an interval of length $h$ would be $$\frac{f(x+h)-f(x)}{h}.$$

But all those expressions speak of a speed over a certain interval. The derivative is interested in speed at a given point, and is therefore calculated as
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
