Derivative is just speed of change?
$f'(x)=f(x+1)-f(x)$
This is, as pointed out by 5xum, not true. Please read his answer to understand the definition of derivative, which states:
$$f'(x) :=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
Let's try to compute this limit for any given point $x$
$$f'(x):=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0} \frac{(x+h)^2-x^2}{h} = \lim_{h\to 0} \frac{2xh+h^2}{h} = \lim_{h\to 0} 2x+h = 2x$$
Note that, for $h=1$, we get indeed your value of $5$, but derivatives are about setting $h \to 0$, not settling for $h=1$
For further intuition, try drawing the parabola $y=x^2$, then draw the tangent line at $x=2$ and also the straight line joining $(2,4)$ and $(3,9)$. See the difference?
Look at how the red line has the slope you predicted (5), and it crosses the parabola at both $x=2$ and $x=3$ while the blue (slope 4) line only touches it at $x=2$, but stays closer to the parabola if you move away a bit. This illustrates that the true slope of the curve is $4$ rather than $5$
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
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$
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.
There is something important missing in your definition. The correct one is:
Derivative is simply an instantaneous speed (or rate) of value change
To see what instantaneous means you can take $\Delta t$ an arbitrary time interval and see what happens when $\Delta t$ gets closer to $0$.
For any real number $\Delta t$ your value change between time $t$ and $t+\Delta t$ is
$$ f(t+\Delta t)-f(t)=(t+\Delta t)^2-t^2=(t^2+2t\Delta t+(\Delta t)^2)-t^2=2t\Delta t+(\Delta t)^2 $$ Now to get speed (or rate) of change you must divide it by the elapsed time between $t$ and $t+\Delta t$, which is $\Delta t$ : $$ \text{rate of change} = \frac{f(t+\Delta t)-f(t)}{\Delta t} = 2t + \Delta t $$ Finally to get the derivative, you must find the instantaneous rate of change, which means that you must find what happens when $\Delta t$ gets closer to zero:
From the previous expression it is clear that when $\Delta t\rightarrow 0$ you have $$ f'(t)=\text{instantaneous rate of change} = \text{rate of change when }\Delta t \rightarrow 0 = 2t + 0 = 2t $$ which is the expected result.
Your main errors were:
to compute the absolute value change $f(x+\Delta t)-f(x)$ and not the rate of change $\frac{f(x+\Delta t)-f(x)}{\Delta t}$
to "ignore" the instantaneous part of the definition, you used $\Delta t = 1$, a finite value instead of considering what happens for the limit case $\Delta t \rightarrow 0$