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I am trying to understand the concept of derivatives more deeply, especially from the point of view of the following equation:

$$\frac{dx^2}{dx} = \lim_{\Delta{x}\to0} \frac{(x+ \Delta x)^2 - x^2}{\Delta{x}} $$

When trying to solve this equation, in the last step we're remaining with the following:

$$\implies \lim_{\Delta x \to 0}[2x + \Delta x] = 2x ~~~\text{(at this particular step how can we put $\Delta x = 0$)}$$

Does putting $\Delta x = 0$ at the second last step is confusing because so far we were assuming change in $x$ tends to $0$ but now we assume it is indeed zero.

Thanks in advance.

killerx
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  • This would be easier to read if you use MathJax (see e.g. http://meta.math.stackexchange.com/questions/5020/tex-latex-mathjax-basic-tutorial-and-quick-reference) to format it. – Henrik supports the community Apr 12 '18 at 18:36
  • Thanks for the pointer, I have tried to fix it better with help of the community – killerx Apr 12 '18 at 18:47
  • You are not putting $\Delta x = 0$. You are taking a limit of $[2x+\Delta x]$ as $\Delta x \to 0$. – Winther Apr 12 '18 at 18:48
  • and what in the case when $$\Delta{x}$$ becomes $$dx$$ – killerx Apr 12 '18 at 18:54
  • If that is confusing you note that it does not matter what you call the variable we take to zero. We might just as well state the definition as $f'(x) = \lim_{h\to 0} \frac{f(x+ h) - f(x)}{h}$. Note that I here called the derivative $f'(x)$ instead of writing $\frac{df(x)}{dx}$. This was just to avoid confusion. The $dx$ in the derivative is just a notation. There is not really any $\Delta x$ becoming $dx$. – Winther Apr 12 '18 at 18:56
  • So it is the limit we're stating as equals? What the result tends to is equal to the limit and that is why we put the equality sign in the example you mentioned above? Is that right? – killerx Apr 12 '18 at 19:04
  • It takes some time to understand that there are some functions for which limit at a point is same as value at that point. Such functions are called continuous and here you can see that $2x+\Delta x$ is a continuous function of $\Delta x$. See more details at https://math.stackexchange.com/a/1822706/72031 – Paramanand Singh Apr 13 '18 at 08:52

1 Answers1

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In the last step you are finding the $$\lim _{\Delta x \to 0} 2x + \Delta x$$

Since $ \Delta x \to 0 $, $2x+\Delta x \to 2x $

Thus $$ \frac {d}{dx}(x^2) = 2x $$

Mohammad Riazi-Kermani
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  • what bothers me is the fact that even though you're using tending sign here, we use actual equality in other places and while solving equations involving derivatives. I feel like the line is blurred, for me between tending and actually putting 0 which bothers me – killerx Apr 12 '18 at 18:47
  • In this case, how can we claim the derivative as equality then? => d/dx(x^2) = 2 x – killerx Apr 12 '18 at 18:51
  • Note that derivative is a limit. Thus if something tends to 0, its limit is 0. If something tends to $2x +\delta x$ while $\delta x$ tends to zero , the limit is $2x.$ – Mohammad Riazi-Kermani Apr 12 '18 at 18:56
  • That makes better sense. Thank you – killerx Apr 12 '18 at 19:04