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We have simple function: $$Y = X^2$$

Writing $X^2$ as: $X^2 = \underbrace{X+X+X+...+X}_{X \text{ times}}$

We can write above equation as:

$$ Y = \underbrace{X+X+X+...+X}_{X \text{ times}}$$

Differentiating with respect to $X$, we get:

$$ \frac{dY}{dX} = \underbrace{1+1+1+1+........+1}_{X \text{ times}}\\ \frac{dY}{dX} =X $$

Since it is known that result must be: $$\frac{dY}{dX}= 2X$$

I just want to know what is wrong with the above derivation.

Kamil Jarosz
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Neeraj
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    $X = 1.5$ times? –  Oct 10 '15 at 09:36
  • what do you want to say..? – Neeraj Oct 10 '15 at 09:37
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    What do you want to say? What is $X + X+ \cdots +X (X$ times) when $X = 1.5$? –  Oct 10 '15 at 09:38
  • I understood your point. But why above derivation is incorrect ? – Neeraj Oct 10 '15 at 09:41
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    Writing $X^2 = X+X+\cdots +X$ only make sense when $X$ is a positive integer. So you cannot differentiate on both side (the right side is just not defined when $X$ is not an integer, so there is no way you can take limit, which is necessary when doing differentiation. –  Oct 10 '15 at 09:43
  • It got your point. But what above result indicate then ? Can we interpret above result ? – Neeraj Oct 10 '15 at 09:49
  • The above example indicate that you have to be very careful when writing $\cdots\cdots$. –  Oct 10 '15 at 09:50
  • Can I say : function can be differentiated only if it is continuous. Writing $X^2 = X+X+....+X$ would discrete(-ize) the above function. – Neeraj Oct 10 '15 at 09:58
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    You forgot that not only do the $X$'s increase, but the number of $X$'s also increase. That means that the derivative isn't simply $1+1+\cdots +1$ $X$ times. – Arthur Oct 10 '15 at 10:05
  • You cannot differentiate both sides of an equality. Because the slopes of LHS and RHS may not be equal. And this is an equality because it is not true for all $X$s. – Aditya Agarwal Oct 10 '15 at 10:08
  • @AdityaAgarwal But it is not an equation. It is an equality of functions. Of course it is true for all $X$, that's how $Y$ was defined, after all. – Arthur Oct 10 '15 at 10:09
  • I agree that it is an equality of functions, but not with that it is true for all $X$s. See, for example, $0.5$. There is nothing as $0.5$ times. @Arthur – Aditya Agarwal Oct 10 '15 at 10:11

2 Answers2

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Let's say we play along, just to see what happens. We have $$ X^2 = X+X+X+\cdots +X\tag{$X$ times} $$ as well as $$ (X+h)^2 = X + X + \cdots + X + h+h+\cdots + h \tag{$X+h$ times} $$ Now, by applying the definition of derivative, we get $$ \frac {dY}{dX} = \lim_{h \to 0}\frac{(X+h)^2 - X^2}{h}\\ = \lim_{h \to 0}\frac{\overbrace{X+X+\cdots+X}^{X+h - X = h\text{ times}} + \overbrace{h + h +\cdots + h}^{X+h\text{ times}}}{h}\\ = \lim_{h \to 0}\left(\frac{\overbrace{X+X+\cdots+X}^{h\text{ times}}}{h} + \frac{\overbrace{h + h +\cdots + h}^{X+h\text{ times}}}{h}\right)\\ = \lim_{h \to 0}\left(\frac{hX}{h} + \frac{(X+h)h}{h}\right)\\ = \lim_{h\to 0} (X + X+h) = 2X $$ I still don't approve (although I do admit it was a fun exercise), but just like pretending $\frac{dY}{dX}$ is an actual fraction that can be simplified gives the correct chain rule, this does give the correct result in the end.

Arthur
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    Thanks Arthur. It is wonderful derivation... – Neeraj Oct 10 '15 at 10:09
  • by op's definition of function, $X$ is restricted to integers. So the function is not continuous. This means IMO you can't apply $f(x+h)$ where h is arbitrarily small real number with your definition of derivative. – Salihcyilmaz Feb 18 '16 at 18:50
  • @Arthur Fighting fire with fire :) – Lev Borisov Feb 18 '16 at 19:02
  • @Salihcyilmaz Did you notice where I said "Let's say we play along"? That was the exact moment I stopped caring about pesky things such as "this doesn't make sense". Did you also notice where I said "I still don't approve"? That's where I say to exactly people like you (and those who might not know that this is wrong) that "You are of course right that this is wrong". – Arthur Feb 18 '16 at 20:13
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$$Y = X^2$$

Wrong: $X^2$ as : $X^2 = X+X+X+...........+X$ $(n\,times, n=2, \implies n X )$ in this case,

Right: $X^2$ as : $X^2 = X * X * X * .....X$ $(n \,times, n=2 ,\implies X^n )$ in this case.

Although multiplication is repeated addition and exponentiation is repeated multiplication, they should not be confused as to their original operative application.

Narasimham
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