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

We all know that the derivative of $x^2$ is 2x. But what if someone proves it to be just x?

## Derivative of x squared

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

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

## A Curious Case

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?

## The 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 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 $x^2$ expand in such a way?

If x is a positive integer:

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

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

$x$ can be written as the sum of its 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.
• $=[x]+\{x\}+x \left({[x]’+\{x\}’ }\right)$
• $=x+x (x’)$
• $=x+x=2x$