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We know that the derivate of the function $f(x)=x^n$ with $n\in \Bbb R$ is:

$$f'(x)=\frac{df}{dx}=nx^{n-1}\tag 1$$

To obtain the proof of the $(1)$ I use a known limit

$$\lim_{u\to 0}\frac{(1+u)^\lambda-1}{u}=\lambda \tag 2$$

In fact

$$\begin{aligned} \frac{df}{dx}&=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{(x+h)^n-x^n}{h}\\ &=\lim_{h\to 0}\frac{x^n\left[\left(\frac{x+h}{x}\right)^n-1\right]}{h}=\lim_{h\to 0}\frac{x^n\left[\left(1+\frac{h}{x}\right)^n-1\right]}{h}\\ &\stackrel{(2)}{=}\lim_{h\to 0}\frac{\frac{x^n}{x}\left[\left(1+\frac{h}{x}\right)^n-1\right]}{\frac{h}{x}}=nx^{n-1} \end{aligned}$$

Is it possible that exist another proof based on a set of algebraic rules, writing the $(x+h)^n-x^n$ in another way?

Sebastiano
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    $x^{n}=e^{n \ln x}$ for $x >0$ so the derivative is $\frac n x e^{n \ln x}=nx^{n-1}$ by Chain Rule. – Kavi Rama Murthy Jan 30 '20 at 08:57
  • In what sense do you mean "without to use the derivative of a product of a function $f(x)$"? In your proof you use the fact that $f'(1) = n$ ("a known limit"). Do you wnat to avoid this? By the way, you need a separate argument for $x = 0$. – Paul Frost Jan 30 '20 at 08:59
  • @KaviRamaMurthy Meanwhile thank you for your attenction for my question. – Sebastiano Jan 30 '20 at 09:05
  • @PaulFrost Thank you also for you. I have edited my question, hoping that it is more clear. – Sebastiano Jan 30 '20 at 09:05
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    My answer to Differentiation using first principles with rational powers shows how you can do this purely algebraically for rational number values of $n.$ To extend this to any real number value of $n$ will require something more mathematically sophisticated (although the algebraic details will not be nearly as involved). Indeed, something more sophisticated is needed just to precisely define something like $2^{\pi},$ and such a precisely defined notion of what it means for an exponent to be a real number needs to be in place before beginning. – Dave L. Renfro Jan 30 '20 at 11:04
  • @DaveL.Renfro Most kind Dave L. Renfro, thank you also for you for another explanation. I have seen your excellent answer that I have upvoted :-). My question is really for my students of an high school. I have in fact (for example) done for a special case $p^n-1^n=(p-1)(1+p+p^2\dotsb +p^{n-1})$. Thank you again. – Sebastiano Jan 30 '20 at 12:33

2 Answers2

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Yes for $n \in \mathbb N$. The binomial theorem gives $$(x+h)^n = \sum_{i=0}^n \binom{n}{i} x^{n-i} h^i = x^n + nx^{n-1}h + \binom{n}{2} x^{n-2} h^2 + \ldots + nx h^{n-1} + h^n .$$ Thus $$(x+h)^n - x^n = \sum_{i=1}^n \binom{n}{i} x^{n-i} h^i = h \sum_{i=1}^n \binom{n}{i} x^{n-i} h^{i-1} , $$ $$\lim_{h \to 0} \frac{(x+h)^n - x^n }{h} = \lim_{h \to 0} \sum_{i=1}^n \binom{n}{i} x^{n-i} h^{i-1} = \sum_{i=1}^n \binom{n}{i} x^{n-i} \lim_{h \to 0} h^{i-1} = n x^{n-1} .$$ We may use this also to treat the case $f(x) =x^r$ for $r = \frac{n}{m} \in \mathbb Q$. We have $f(x) = (x^n)^{1/m} = \sqrt[m]{x^n}$. With $g(x) = x^m$ the derivative of $g^{-1}(y) = y^{1/m} = \sqrt[m]{y}$ is $$(g^{-1})'(y) = \frac{1}{g'(g^{-1}(y))} = \frac{1}{g'(y^{1/m})} = \frac{1}{m y^{(m-1)/m}} = \frac{1}{m}y^{(1/m) -1} .$$ Thus by the chain rule $$f'(x) = g'(x^n) nx^{n-1} = \frac{1}{m}x^{n((1/m) -1)}nx^{n-1} = \frac{n}{m}x^{(n/m) -n + n-1} = rx^{r-1} .$$

Paul Frost
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    Right, it only works for $n \in \mathbb N$. For $n \notin \mathbb N$ you must somehow define $x^n$, and the most popular definition is $x^n = e^{n\ln x}$. See Kavi Rama Murthy's comment. – Paul Frost Jan 30 '20 at 09:28
  • Thank you very...very much. Please, can you explain, after, the step in the previous my comment editing your answer? – Sebastiano Jan 30 '20 at 09:33
  • What precisely should I explain? The step after the binomial theorem? – Paul Frost Jan 30 '20 at 11:31
  • You have read my thoughts :-). Ah now I have understood: $(x+h)^n - x^n =nx^{n-1}h + \binom{n}{2} x^{n-2} h^2 + \ldots + nx h^{n-1} + h^n$. Thank you so much. – Sebastiano Jan 30 '20 at 12:36
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If you are allowed the derivative of a product, by induction

$$(x^1)'=1\cdot x^0$$ and $$(x^{n-1})'=(n-1)x^{n-2}\implies(x^n)'=(x\cdot x^{n-1})'=x^{n-1}+x\cdot(n-1)x^{n-2}=n x^{n-1}.$$

In fact we have enough with a special form of the product rule, which is straightforward to prove:

$$\frac{(x+h)f(x+h)-xf(x)}h=x\frac{f(x+h)-f(x)}h+f(x+h)\implies (xf(x))'=f(x)+xf'(x).$$

Alternatively

$$\frac{(x+h)^n-x^n}h=\frac{x+h-x}h((x+h)^{n-1}+(x+h)^{n-2}x+\cdots x^{n-1})\to x^{n-1}+x^{n-1}+\cdots x^{n-1}.$$

There are $n$ terms.

  • I approved very much also your answer and thank you again. Today there are the matters of exam of Italian State for the students of high school. – Sebastiano Jan 30 '20 at 12:53
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    @Sebastiano: I guess that my second method is what you were looking for. –  Jan 30 '20 at 12:56
  • Last method is perfect! :-) Very nice. But just only for my humble sincerity I have given, before, the check mark to Paul Frost. For me the respect and the sincerity are importants. – Sebastiano Jan 30 '20 at 13:01
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    @Sebastiano: it is a common behavior on this site to rush on the first answer (even a wrong one, but that's another story). Never mind. –  Jan 30 '20 at 13:02
  • I have not forgotten your advice as in the site TeX.SE. – Sebastiano Jan 30 '20 at 13:04