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..


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

Now let

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


$ 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.

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

$=x+x (x’)$


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

Multiplication is not repeated Addition.

  • Greatest Integer Function
Default image
Gaurav Tiwari
Gaurav Tiwari is a professional graphic & web designer from New Delhi, India. is his personal space where he writes on blogging, digital marketing, content writing, learning and business growth. Gaurav has contributed in developing more than 325 brands worldwide and while you are reading this, he's busy building a couple more.

Newsletter Updates

Enter your email address below to subscribe to our newsletter


  1. 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.

  2. 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.

  3. Its obvious..The fault is in the beginning itself..Why you are making very absurd assumption.
    You cannot write $x^2=x+x+x+\ldots$, but you can write $x^2=x+x$.
    How can you say
    “However, suppose we write $x^2$ as the sum of x ‘s written up x times..” If it your assumption, then it is not $x^2$..actually it is for $x^x$.
    Got it!

  4. I think better change to:
    Derivative of x squared is 2 ? Where is the fallacy?
    Yes x^x means x multiplied to itself x number of times
    and x^2 means x multiplied to itself, i.e x times x or X x X.
    But you say x^2 means x multiplied to itself x number of times.

Leave a Reply