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Derivative of $x+x+x+x \ldots$ $x$ times wrt $x$ is $2x$. Why?

I know that derivative of $x^2 = 2x$. But if someone asks $1+1+1 \ldots x$ times $= x$, not $2x$, what will be the clear explanation for that? Like if I had to explain it to someone clearly, how would I go about it?

Robert Shore
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Anne
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  • $x+x+x...$ where there are $x$ elements is $x^2$, whose derivative is $2 x$, valid solely for $x$ a positive integer. – David G. Stork Aug 19 '23 at 17:30
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    Can you tell us the value of x+x+x+x... x times when $x = 5/2$? – mathcounterexamples.net Aug 19 '23 at 17:30
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    The fundamental issue is that the number of times you're adding $x$ to itself also varies with $x$, and merely adding $1$ to itself $x$ times doesn't account for that additional variance. – Robert Shore Aug 19 '23 at 17:30
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    @mathcounterexamples.net great point. To add slightly: the problem with differentiating such a function is that it has no natural extension to $\mathbb{R}$ which means we can't do calculus with it. You'll probably want to extend it to $x^2$, but in that case the answer is of course $2x$ as you can prove. – Charles Hudgins Aug 19 '23 at 17:41
  • the product rule says the derivative of $x \cdot x$ is $1 \cdot x + x \cdot 1$ – Will Jagy Aug 19 '23 at 17:41
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    This has been covered many times already. Define $$f:(x,y)\mapsto \underbrace{x+\dots+x}_{y\text{ times}}$$ and let $g: x\mapsto (x,x)$. The function you are proposing to differentiate is $f(g(x))$, where we ignore what it actually means to define $f$. Proceeding heuristically, we get $f'(g(x))g'(x) = \partial_1 f(x,x)+\partial_2 f(x,x)$. Now, it is obvious $\partial_1 f(x,x) = x$. The difficulty is in assigning a value to $\partial_2 f(x,x)$, but it is reasonable to claim $\partial_2 f(x,x) = x$, which yields the "correct" answer, as expected. – Andrew Aug 19 '23 at 17:46
  • Of course, the function $f$ is really $f:(x,y)\mapsto xy$ since this is the generalization of $y$ times to non-integer values of $y$, in which case the answer is without a doubt $2x$. The responses by david and mathcounterexamples fail to capture the real issue why your argument fails, namely that you are not using the chain rule. – Andrew Aug 19 '23 at 17:48
  • See this MSE answer. For an ASCII-text diagram explanation, scroll down to "Dave Renfro posed the following" here. – Dave L. Renfro Aug 19 '23 at 17:54
  • Right, so it seems the problem is actually in writing the expression as x^2. Since, we can't really expand it unless x is a positive integer. Thank you everyone. – Anne Aug 19 '23 at 17:56
  • @Andrew I strongly disagree. Before everything in math, you have to rely on well-thought definitions. Here, the definitions are not only fuzzy but inexistent. So first try to define properly what you speak about and only then make extrapolations. The real issue is that $x + \dots + x$ $x$ times doesn't make sense and that has to be said. – mathcounterexamples.net Aug 19 '23 at 19:40
  • @mathcounterexamples.net No, the real issue is that $f:(x,y)\mapsto xy, g:x\mapsto (x,x)$ and to differentiate $h(x) = f(g(x))$, the op forgot to apply chain rule. Indeed, op only computed $\partial_1 f(g(x))$ whereas $h'(x) = \partial_1 f(g(x))+\partial_2f(g(x))$ – Andrew Aug 19 '23 at 20:04
  • Besides, what one can always do is argue as follows. Look at $x^2 = x\cdot x$, which evaluated at any integer $n$ is equal to $x+\dots+x$, $n$ times. It is then OK to differentiate this function term by term and arriving at the answer that $d/dx ,x^2$ equals $n$ when evaluated at integers $n$. Your objection that the OP's function is not "defined" is evidently not the issue. The problem can always be restated in a way that makes sense and the issue is that $n$ is a function of $x$. – Andrew Aug 19 '23 at 20:12

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First, consider the definition of multiplication 3 + 3 3 times is 9 because 3 + 3 + 3 = 9. 2 + 2 two times is four. x + x x times should be $x^2$ following this pattern. To prove this, look at the pattern closer. 3 + 3 3 times is 3 + 3 + 3, or 3 * 3, or $3^2$, or 9. x + x x times is x * x, or $x^2$. The derivative of $x^2$ is 2x. When solving any math problems, always try to get the problem to its simplest form before even attempting to solve it, as shown in this problem.

Berny
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  • No, the derivative of $x^2$ is $2x$ from the definition of the derivative. Multiplication is not always repeated addition. – Accelerator Aug 20 '23 at 21:29