Proof that 2=1 using differentiation

Question

We can express a square number as the repeated addition of that number in this manner:

$1^2 = 1$

$2^2 = 2 + 2$

$3^2 = 3 + 3 + 3$

Generalising this, we get:

$x^2 = x + x + x...$ $x$ $times$

If we differentiate with respect to $x$ on both sides, we get:

$2x = 1 + 1 + 1...$ $x$ $times$

$2x = x$

$2 = 1$

This is obviously wrong. What's the mistake in my proof?

Answer 1

Your expression of repeated summation is:

$$x^2=x\cdot x = \underbrace{x+x+x+\cdots}_{x\,\text{times}}$$

Underlined which $x$ you forgot to differentiate over:

$$x\cdot \underline{x} = \underbrace{x+x+x+\cdots}_{\underline{x}\,\text{times}}$$

Of course, it gets worse. Differentiation is defined for continuously varying arguments (e.g. real numbers). Repeated addition only works for natural numbers. So this cannot work in this form, until you move on from repeated summation definition.

Answer 2

Because when you differentiate, you forgot to deal with the '$x$ times' expression, which is non-constant.

Answer 3

The formula only holds for natural values of $x$. Hence you can't differentiate.