Derivative of $x^2$ This seems too easy, but here's the question:
$x^2$ is $x + x + ...+ x$ (with $x$ terms). Its
derivative is $1 + 1 + ... + 1$ (also $x$ terms). So the derivative
of $x^2$ seems to be $x$.
And another expression: we know that if $y = nx$, then $y' = n$, so that if $y = x * x$ then $y' = x$.
But we know by formula that if $y = x^2$, then $y' = 2x$
So, how to prove $y' = x$ is wrong ?
Thanks
 A: For the first "paradox", you simply found a way to make it look like the number of terms in the sum is constant.  In reality, the number of terms is increasing with $x$.  Imagine if you wrote $x=1+1+...+1$ ($x$ terms), then differentiated to get that $x'=0$.  But you forgot to take the variation in length of the sum into account.  This version is perhaps more transparent because you no longer have the illusion that you took $x$ into account already.  
For the second paradox, that formula $y=nx \Rightarrow y'=n$ very explicitly depends on the fact that $n$ is constant with respect to $x$.  Otherwise the fact is simply wrong.  
A: The answer lies in the subtle difference between
$$f(x) = x^2$$
and
$$f(x) = x + x + x + \cdots + x (x\ \mathrm{times}).$$
What is this? Well, the first expression is used to define a function on real numbers. The second expression refers to a function on natural numbers. Namely, "x times" only makes sense for $x$ a natural number. What does it mean to add $x$ to itself $\frac{1}{2}$ of a time? It doesn't make sense.
The idea of multiplication as a "repeated addition" doesn't work when you get to multiplying non-natural numbers. It's better to think of a real product $ab$ not as "$a$ added to itself $b$ times", but rather like "$a$ scaled by the scale factor $b$". A scaling operation can be varied in intensity continuously; a process of repetition cannot.
So, not being a function of real numbers, you cannot take its derivative.
A: Below is my reply to this in an old sci.math post.
Zachary Turner  wrote on 26 Jul 2002:

Let D = d/dx = derivative wrt x. Then
D[x^2] = D[x  +   x  + ... +   x  (x times)]
       = D[x] + D[x] + ... + D[x] (x times)
       =   1  +   1  + ... +   1  (x times)
       =   x


Notice that an obvious analogous fallacious argument proves both
  D[x f(x)]  =  Df(x) (x  times) = x Df(x)

  D[x f(x)]  =   Dx (f(x) times) = f(x), via  Dx = 1

vs.  the correct result: their sum  f(x) + x Df(x)
as given by the Leibniz product rule (= chain rule for times).
The error arises from overlooking the dependence upon x in both
arguments of the product  x * f(x)  when applying the chain rule.
The source of the error becomes clearer if we consider a
discrete analog. This will also eliminate any tangential
concerns on the meaning of "(x times)" for non-integer x.
Namely, we consider the shift operator  S: n $\to$ n+1  on polynomials  p(n)
with integer coef's, where  S p(n) = p(n+1). Here's a similar fallacy
  S[n^2] =  S[n  +   n  + ... +   n  (n times)]
         =  S[n] + S[n] + ... + S[n] (n times)
         =  n+1  + n+1  + ... + n+1  (n times)
         = (n+1)n

But correct is $\rm\: S[n^2] = (n\!+\!1)^2.\:$ Here the "product rule" is
simply  S[fg] = S[f] S[g], not  S[f] g,  as above.
The fallacy actually boils down to operator non-commutativity.
On the space of functions f(x), consider "x" as the linear
operator of multiplication by  x, so  x: f(x) $\to$ x f(x). Then
the linear operators  D  and  x  generate an operator algebra
of polynomials  p(x,D) in NON-commutative indeterminates x,D
since we have
  (Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so  Dx = xD+1

  (Sn)[f] = S[nf] = (n+1)S[f], so  Sn = (n+1)S ≠ nS

This viewpoint reveals the error simply as mistakenly
assuming commutativity of the operators  x,D  or  n,S.
Perhaps something to ponder on boring commutes !
A: Here is a rough argument as to why this is wrong. Add 1 to $x$ - you get your 1+1+1 ... ($x$ terms, total $x$) as the difference, but you also get an additional term of value approximately $x$ [strictly $x+1$ - because you are adding more $x$s] so the derivative is $2x$.
A: Everyone is writing down his favorite version of the explanation.  Try this:  Define
$$
f(x,y) := x+x+\dots+x\qquad\text{($y$ terms)}
$$
Then "compute" that
$$\begin{align}
\frac{\partial f}{\partial x}(x,y) &= 1+1+\dots+1\qquad\text{($y$ terms)}
\\
&= y,
\\
\frac{\partial f}{\partial y}(x,y) &= x+x+\dots+x\qquad\text{($1$ terms)}
\\
&= x
\end{align}$$
Now suppose $x=z, y=z$, and so, according to the chain rule for partial derivatives:
$$\begin{align}
\frac{d}{dz}\big[z+z+\dots+z&\qquad\text{($z$ terms)}\big]
=\frac{d }{dz}f(z,z)
\\
&=\frac{\partial f}{\partial x}(z,z)\frac{dx}{dz}+
\frac{\partial f}{\partial y}(z,z)\frac{dy}{dz}
\\
&= z + z = 2z
\end{align}$$
A: OK. Before I learnt rigour, I would normally respond to such questions by an answer like the number of x terms is varying. This answer might give you an idea about what is going wrong. However, I really prefer this answer, because the other answer ignores the fact that you are adding x terms. 
Addition is a binary operation. You can use this fact along with the fact that addition is associative to evaluate the sum of three numbers as: a+(b+c) or (a+b)+c. You can use this to add 4 numbers or 5 numbers ....
Now at the beginning of your argument you said $x^2=\sum\limits_{n=1}^{x}x$ which does not make sense for all x except positive  integers.
Summation operators $\sum\limits_{n=1}^{k}a_k$ are defined recursively. This definition is valid for positive integers k. From this definition it becomes possible to PROVE that :
$(\sum\limits_{n=1}^{k}f_k(x))'=\sum\limits_{n=1}^{k}f'_k(x)$
$\sum\limits_{n=1}^{k}a=ka$
However, this defintino does not allow all real values of x. Thus, when you are taking about $x^2=\sum\limits_{n=1}^{x}x$ you can not be using the standard definition of summation. In this case you have to tell us the definition of summation that you are using. For your argument to work, your definition of $\sum\limits_{n=1}^{x}f_n(x)$  should satisfy the following (you should PROVE that your definition satisfies the following):
1) It should be defined for all real values of x.
2) It should satisfy: $kx=\sum\limits_{n=1}^{x}k$
3) It should satisfy: $(\sum\limits_{n=1}^{x}f'_n(x))=(\sum\limits_{n=1}^{x}f_n(x))'$
However you can't just assume properties of $\sum\limits_{n=1}^{x}f_n(x)$ because the notation is similar to $\sum\limits_{n=1}^{k}a_n)$  !   
