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

### Start here

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?

Well!

I got following faults:

•One should note that $x^2$ is always a positive quantity. Whereas x can be either positive or negative.

But if we sum negatives any number of times, we get a negative number only.

Thus, x²=x+x+….x (x times) [this expression is proven wrong below] or $nx=x+x+....+x$ (n times)

•The number of x’s being added is not a constant. Not only is x changing; the number of x’s is also changing. And one must note that nx=x+x+x+…..+x (x written n times)

But as we know that nx=xn

Thus one may think that xn=n+n+n+…..+n (n written x times).

One who thinks the second above, is wrong. A number / variable added a variable (number) of times doesn’t follow the expression xn=n+n+….+n (x times) but it follows xn=x+x+….+x (n times). I mean to say that a variable/constant can’t be added a variable number of times.

I’m a student and I wrote – what I knew. So there were many incompletenesses in my post. I got these excellent suggestions and updates on this post:

$x^2 = \displaystyle {\left( {x+x+\ldots +x} \right)_{W(x) \, times}} + F(x) x$

where W(x) is the whole number part of x and F(x) is the fractional part of x.

So

x² = xW(x)+xF(x)

d/dx [x²]= d/dx[xW(x) +x F (x)]

(differentiation by part)

= x’•W(x)+x • W’(x) + x’•F(x) + x•F’(x)

since x’ =d/dxx=1 and with W(x) being a constant

d/dx[x²]

= W(x)+F(x)+x {W’(x)+F’(x)} As, W(x)+F(x)=x & W’(x)+F’(x)=1 so d/dxx² = x+x=2x

[Credit: goofindoo]

### Update-IV

This may be an excellent additional reading for readers, which was published on one of the my favorite blogs – Republic of Mathematics:

Practical Applied Mathematics: Wherein Multiplication is treated as repeated addition

updated: 02/14/2011 11:38 am.

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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 adition: 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 says:

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.

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