Why is the derivative of $x^2$ not $2x+1$? If the derivative is the change of the function at each step, it could be expressed as:
$$f(x)+f'(x)=f(x+1)$$
Therefore if $f(x)=c$
$$c+f'(x)=c \implies f'(x)=0$$
This is also correct for $f(x)=cx$
$$cx+f'(x)=c(x+1) \implies f'(x)=c$$
However, it doesn't work for $f(x)=x^2$
$$x^2+f'(x)=(x+1)^2=x^2+2x+1 \implies f'(x)=2x+1$$
Where am I wrong in my reasoning?
 A: You are confusing discrete functions, that is, functions which are only defined for integers or natural numbers, with continuous functions on the real numbers.
The derivative of a function like $f(x)=x$ on all the real numbers is NOT defined as the change of the function at each step because there are no such things as steps. 
If, however, you only consider discrete functions and actually call the increment of the function at each step its derivative, you are right when you say that the derivative of $f(n)=n^2$ is $2n+1$.
A: Where are you wrong?
In your first equallity:
$x^2 + f′(x)=(x+1)^2$
You can't suppose that when you sum a derivate you will sum one to the argument of the function...
A: The thing you take for a derivative is
$$
\Delta_1f(x)=f(x+1)-f(x).
$$
Similarly, you can define an analogous thing for different step sizes $h\neq0$:
$$
\Delta_hf(x)=\frac{f(x+h)-f(x)}{h}.
$$
The derivative is defined as the limit of these difference quotients as $h\to0$:
$$
f'(x)=\lim_{h\to0}\Delta_hf(x).
$$
Let's consider your concrete example, $f(x)=x^2$.
For it we can calculate that $\Delta_hf(x)=2x+h$, so that $\Delta_hf(x)\to2x$ as $h\to0$.
In short, the main point of this answer is that typically $\Delta_1f(x)\neq f'(x)$.
A: The rate of change is itself changing. In order to get from $f(a)=a^2$ to $f(a+1)=a^2+(2a+1)$, earlier (closer to $x=a$) the $x^2$ function is changing more slowly than when you get closer to $x=a+1$.  Overall the average rate of change is $2a+1$, but instantaneously, at $a$ the rate of change is $2a$, while at $a+1$ the rate of change is $2a+2$.
