Maybe it is a soft question, I can't find any information about this. How people understood that we can use derivative shortcuts instead of calculating limits and how did they derive them?

By shortcuts I mean rules like $(x^n)^{'} = nx^{n-1}$. I find them somehow intuitive, like if we have a linear function, its derivative should be constant, etc. But I can't figure out how one can derive those shortcuts. Any references, links will be appreciated.

P.S. I am also interested in historical part who found them. Were they proved at once by someone? Did Newton or Leibniz know these shortcuts?

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    $\begingroup$ They were found by applying $f (x+\delta x) - f(x)=f'(x)$ to a wide range of powers and then proving that it holds for all $n$. $\endgroup$ – Mohammad Zuhair Khan Dec 28 '17 at 21:16
  • $\begingroup$ Like any other shortcut ever in mathematics, it can be used because the long way of doing it has already been done and proven, so instead of doing it again from scratch each time, we just remember the result and apply it. Other examples include a lot of formulas like the binomial theorem and the quadratic formula. $\endgroup$ – Arthur Dec 28 '17 at 21:19
  • $\begingroup$ @MohammadZuhairKhan, did Newton know them? Or is there a name of the mathematician, who found them? I mean explicitly shortcuts. $\endgroup$ – Turkhan Badalov Dec 28 '17 at 21:22
  • $\begingroup$ They were probably found by Newton or Leibniz but the $2$ answers are way simpler and relevant. $\endgroup$ – Mohammad Zuhair Khan Dec 28 '17 at 21:25
  • $\begingroup$ If you are still interested in the history part, you can ask specifically for that on hsm.stackexchange.com. $\endgroup$ – Torsten Schoeneberg Jan 6 '18 at 3:25

You can compute directly from the definition of the derivative. For your power rule example, there are some ways to do this, none of which are L'Hopital's rule.

$$ \lim_{h\to 0}\frac{1}{h}((x+h)^n-x^n)\\ \stackrel{\text{binomial formula}}{=} \lim_{h\to 0}\frac{1}{h}\left(x^n-x^n+nx^{n-1}h+h^2\sum_{k=0}^n{n \choose k}x^{n-k}h^{k-2}\right)\\ =nx^{n-1}+\lim_{h\to 0}h\sum_{k=0}^n{n \choose k}x^{n-k}h^{k-2}=nx^{n-1} $$ the derivative of $f(x)=x^n$ at $x$.


If you can take limits on tangents to the line $x^2$ as you approach some point on the line then you are virtually there.

Rather than just calculate limits for individual functions such as $x^2$ to get $\dfrac{dx^2}{dx}=2x$ all you need do is repeat the same for general forms such as $x^n$ which yields the general form of the derivative $\dfrac{dx^n}{dx}=nx^{n-1}$.

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    $\begingroup$ After noticing that the formula seems to be $(x^n)' = nx^{n-1}$ one can use induction over $n$ to prove it (one needs the formula for derivative of a product). $\endgroup$ – md2perpe Dec 30 '17 at 15:57
  • $\begingroup$ @md2perpe good point; can I add to the answer? $\endgroup$ – samerivertwice Dec 30 '17 at 15:59
  • $\begingroup$ Of course you can. $\endgroup$ – md2perpe Dec 30 '17 at 16:02

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