# Derivative of x squared is 2x or x ? Where is the fallacy?

As we know, the derivative of x squared, i.e, differentiation of $x^2$ , with respect to $x$ , is $2x$.

i.e., $\dfrac{d}{dx} x^2 = 2x$

However, suppose we write $x^2$ as the sum of $x$ ‘s written up $x$ times..

i.e.,

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

Now let

$f(x) = \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}}$

then,

$f'(x) = \dfrac{d}{dx} \left( \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}} \right)$

$f'(x)=\displaystyle {\underbrace {\dfrac{d}{dx} x + \dfrac{d}{dx} x + \ldots + \dfrac{d}{dx} x}_{x \ times}}$

$f'(x)=\displaystyle {\underbrace {1 + 1 + \ldots + 1 }_{x \ times}}$

$f'(x) = x$

This argument appears to show that the derivative of $x^2$ , with respect to $x$, is actually x, not 2x..

Where is the error?

Error: $x^2$ will equal to $\displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}}$ only when $x$ is a positive integer (i.e., $x \in \mathbb{Z}^+$. But for the differentiation, we define a function as the function of a real variable. Therefore, as $x$ is a real number, there arises a domain $\mathbb{R}- \mathbb{Z}^+$ where the statement $x^2= \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}}$ fails.

And since, the expansion  $x^2 \neq \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}}$  for $x \in \mathbb{R}$ , the respective differentiations will not be equal to each other.

## Then how can $x^2$ expanded in such a way?

If x is a positive integer:

$x^2= \displaystyle {\underbrace {x+x+x+ \ldots +x}_{x \ times}}$.

But when when x is an arbitrary real number >0, then

$x$ can be written as the sum of it’s greatest integer function [x] and fractional part function {x}.

Therefore, $x^2 = [x] \cdot x + {x} \cdot x$

$x^2 = \displaystyle {\left( {x+x+\ldots +x} \right)_{[x] \, \mathrm{times}}} + x \cdot {x}$

So, we can now correct the fallacy by changing the solution steps to:

$x^2 = x[x]+x\{x\}$

$d/dx {[x²]}= d/dx \left( {x[x] +x \{x\} }\right)$

(differentiation by part)

$= 1\cdot [x]+x \cdot [x]’+ 1\cdot \{x\} + x \cdot \{x\}’$

since $d/dx (x)=x’=1$ and [x]’ & {x}’ represent differentiation of each with respect to x.

$=[x]+\{x\}+x \left({[x]’+\{x\}’ }\right)$

$=x+x (x’)$

$=x+x=2x$