# Where is the flaw in this “proof” that 1=2? (Derivative of repeated addition)

Consider the following:

• $1 = 1^2$
• $2 + 2 = 2^2$
• $3 + 3 + 3 = 3^2$

Therefore,

• $\underbrace{x + x + x + \ldots + x}_{x \textrm{ times}}= x^2$

Take the derivative of lhs and rhs and we get:

• $\underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = 2x$

Which simplifies to:

• $x = 2x$

and hence

• $1 = 2$.

Clearly something is wrong but I am unable pinpoint my mistake.

To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + \dots + u$, $v$ times. Then $x^2 = F(x,x)$. Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true. In the fallacious example, the problem is essentially that the $F_v$ term has been omitted. In some sense, one has forgotten to differentiate the operation “$x$ times” with respect to $x$! Of course, the notation makes this easier to do.