# 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?

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.

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

$=x+x (x’)$

$=x+x=2x$

- Yesmanapple sent his view on this article. Have a look.

*wnoise suggested this link:*

*Multiplication is not repeated Addition.*

- Greatest Integer Function

## 5 comments add your comment

You simply failed to take account of the fact that not only the value of x changes, but also the size of the set itself, which you didn’t. In reaction to the second reply:

x² = xW(x)+xF(x) Why not just write x² = xW(x) = x*x ? Then you can differentiate this by parts as well.

And why isn’t multiplication repeated addition? The blog only says it isn’t, without explaining why. As far as I know, multiplication is repeated addition. This fact is very useful if you need to multiply long numbers, like 1,345,843 *3,464,901, in your head or with paper.

Hi! Thanks for your comment. $ x^2 =x+x+x+ldots +x$ is true, if and only if x is a positive integer.

But x*x is as same as:

x*x =x*([x]+{x})

where [x] is integer part of x and {x} is fractional part of x. This post is very old and it need to be edited since I had used W(x) and F(x) for [x] and {x} respectively.

Regarding, multiplication is not repeated addition: How can you explain— $ {5.74}^2$, or $ {-4}^2$ as addition? One can’t add any number fractional number or negative number of times.

4^2 = 4 * 4 = 4 + 4 + 4 + 4

$4$ is a fixed positive integer. You can add things upto 4 times, but not all $ x in mathbb{R}$. Differentiation, here, is defined on real numbers.

I FOUND IT VERY VERY VERY EASY QUESTION BECAUSE I ALREADY KNEW IT