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 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.

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}. (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’)

=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

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.

Navigation

Meet the editor

I am Gaurav Tiwari, a freelance web designer and blogger. With an experience of over five years with WordPress, PHP and CSS3, I am capable of doing almost anything related to WordPress CMS. Beyond that, I am a mathematics graduate & a civil service aspirant. Learn more about me.

Subscribe to Blog via Email

Recent Posts

- The disadvantages of Free Web Hosting
- How to fix your damaged pendrives & memory cards?
- Best Hotel Management Colleges in India
- Best Mass Communication Colleges in India
- Affordable Smartphones – What to look for?
- Be Smarter and Nicer by Reading Literature
- How to Start Internet Marketing with a single website?
- Best Commerce Colleges in India
- Best Arts Colleges in India
- The 10 most obvious reasons for using responsive design in 2015

Recent Comments

- The disadvantages of Free Web Hosting | Gaurav Tiwari on How to Start Internet Marketing with a single website?
- Jeff Guynn on Get countless Backlinks with Gravatar Profiles
- Best Mass Communication Colleges in India on Best Arts Colleges in India
- Paul Wandason on Be Smarter and Nicer by Reading Literature
- Rahul Chopra on Best Commerce Colleges in India

© Gaurav Tiwari

Built with Code & Coffee. Powered by Love.

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.

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!

Hmmm. Weird comment. $x+x$ is $2x$ not $x^2$.

And, $x^x$ means $x$ multiplied to itself $x$ number of times.

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.

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

.

Please read the article once again and http://www.maa.org/external_archive/devlin/devlin_01_11.html <– this one too.

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