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

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

You can either start a new conversation or continue an existing one.Please don't use this comment form just to build backlinks. If your comment is not good enough and if in some ways you are trying to just build links — your comment will be deleted. Use this form to build a better and cleaner commenting ecosystem. Students are welcome to ask for help, freebies and more. Your email will not be published or used for any purposes.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.

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.

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.