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

February 6, 2011/9/0
Home / Derivative of x squared is 2x or x ? Where is the fallacy?

As we know that the derivative 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 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}}


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


can be written as the sum of it’s greatest integer function [x] and fractional part function {x}.  (See this video for more details.)

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.




  • wnoise suggested this link:

Multiplication is not repeated Addition.

  • Greatest Integer Function

(last updated on 13th December 2013, 12:45:17 PM IST)


Excellent Reader, Good Orator and Bad Writer. Talk about WordPress, PHP and Mathematics. I develop & design web applications that are visually brilliant and mass-friendly. I can also design stunning graphics to support my web-apps & amazing logos to amaze people. I have worked with several companies & individuals to deliver satisfactory results.
Comments (9)
  • Luitzen Hietkamp / November 4, 2011 / Reply

    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.

  • Gaurav Tiwari / November 4, 2011 / Reply

    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.

    • j / April 2, 2012 / Reply

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

      • Gaurav Tiwari / April 2, 2012 / Reply

        $4$ 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.

  • Raja / December 13, 2013 / Reply

    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!

    • (Author) Gaurav Tiwari / December 13, 2013 / Reply

      Hmmm. Weird comment. $x+x$ is $2x$ not $x^2$.
      And, $x^x$ means $x$ multiplied to itself $x$ number of times.

  • Raja / December 13, 2013 / Reply

    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.

    • (Author) Gaurav Tiwari / December 13, 2013 / Reply

      No. I have written, $x^2$ as the sum of x‘s written up x-times, not that

      x^2 means x multiplied to itself x number of times

      Please read the article once again and <– this one too.

    • Ankur / July 15, 2014 / Reply

      do u even know the ‘M’ of mathematics? :p

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