Just got confused with what my friend asked (paradox and fake proofs). Take $x^2=x+x+x+\cdots$ ($x$ times).
Now differentiating both sides wrt $x$, we get:
$$2x=x.$$
This means $x=0$ or $2=1$.
How? Where did I go wrong?
 A: What you are not considering is that while you are taking derivatives of the function on the right hand side with respect to x, the number of components is not fixed. For your example, what you are doing is basically looking at the function
$$
g(x,n) = xn
$$
and taking the derivative with respect to $x$ and evaluating the partial derivative at $n=x$, i.e. $g_x(x,n)|_{n=x}=n|_{n=x}=x$. 
You would need to look at the total derivative of $g$ at $n=x$, i.e.
$$
dg = g_{x}|_{n=x} + g_n|_{n=x} = 2x.
$$
A: The problem, in short, is that the equation $$x^2 = \underbrace{x+\cdots+x}_x$$ only holds when $x$ is a natural number. 

To truly see why this is a problem, however, you will need to already understand universal quantification and lambda abstraction. Perhaps look up some YouTube videos about these concepts, or just browse on the internet for awhile until you find good explanations.
Explicitly, what we know is:
$$\left(\mathop{\forall}_{x \in \mathbb{N}}\right)\;x^2 = \underbrace{x+\cdots+x}_x$$
In function notation:
$$\left(\mathop{\lambda}_{x \in \mathbb{N}} x^2\right) = \left(\mathop{\lambda}_{x \in \mathbb{N}} \underbrace{x+\cdots+x}_x\right)$$
So if by $D$ we mean differentiation of functions $\mathbb{R} \rightarrow \mathbb{R}$, we can't use the above equation to deduce 
$$D\left(\mathop{\lambda}_{x \in \mathbb{N}} x^2\right) = D\left(\mathop{\lambda}_{x \in \mathbb{N}} \underbrace{x+\cdots+x}_x\right),$$
because neither left- nor right-hand-sides are well-defined.
A: My Explanation:
Looking at the definition of a derivative:
$$(x^2)'=\lim_{h \to 0} \frac{(x+h)^2-x^2}{h}$$
$x+h\notin\mathbb{N}$, if $x\in\mathbb{N}$.
$x^2 = \underbrace{x+\cdots+x}_x$ is true only when $x\in\mathbb{N}$.
A: The idea to differentiate a function in order to find the zeros of the function is a big mistake because the zeros of the function are different from the zeros of it's derivative.
A elementary example : Solve $x^3-x=0$ for $x$
$f(x)=x^3-x$ . The roots of $f(x)=0$ are $-1\:,\:0\:,\:1$
$f'(x)=3x^2-1$ . The roots of $f'(x)=0$ are $-\frac{1}{\sqrt{3}}\:,\:\frac{1}{\sqrt{3}}$
Solving $f'(x)=0$ for $x$ doesn't give the roots of $f(x)=0$
That is the same in case of an equation made of the equality of two different functions : $f(x)=g(x)$
Of course, the functions $f$ and $g$ are different and we have $f(x)=g(x)$ only for a few values of $x$ that we call the roots of the equation. Except for the roots we have $f(x)\neq g(x)$ and thus $f'(x)\neq g'(x)$. 
Now, if we consider the equation $f'(x)=g'(x)$ this equality holds only for a few values and for all other values $f'(x)\neq g'(x)$. There is not raison at all for the roots of the derivated equation be the same as the roots of the original equation.
