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I just cannot prove that $$\frac {d}{dx} {x^n} = n x^{n-1}$$ for $ n \in \Bbb R$.

For $n \in \Bbb{N}$, I can use the definition of a derivative :

$$\frac {d}{dx}x^n = \lim_{h \rightarrow 0} \frac{(x+h)^n - x^n}{h}$$

Now applying "Binomial Expansion" for $\displaystyle (x+h)^n=\sum_{i=0}^{n}{n \choose i }x^{n-i}h^i$ and expanding, the $x^n$ term in the numerator cancels out and the $h$ from denominator divides the entire remaining expression . Taking limit $h$ tending to $0$ gives the required result.

I have been taught that the derivative result holds for all real $n$. But I am not aware of any "formula" which can allow me to expand a binomial expression with real index. I do know about the Taylor Expansion, but if I remember correctly, it utilises the very derivative that I am trying to find.

How can I proceed ?

cqfd
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    How about this formula: $x^n=\exp(n\ln x)$? – Angina Seng Feb 08 '19 at 07:34
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    @LordSharktheUnknown But this will hold only for $x > 0$. –  Feb 08 '19 at 07:38
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    @Nirbhay: if you are interested in non-integer $n$, then there are difficulties in considering $x^n$ for $x\lt0$. – robjohn Feb 08 '19 at 07:40
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    @robjohn Yeah got it. Thanks. –  Feb 08 '19 at 07:41
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    @taritgoswami I know all the "rules" and stuff. But will the product rule be useful ? –  Feb 08 '19 at 07:42
  • @Nirbhay If you want to use induction on $n$ then yes, but as you are saying about $n\in \mathbb{R}$ it is not useful, as we can't use induction in $\mathbb{R}$. The point I made is the rules students taught later may not be known to you(as it seemed you are new to calculus). It doesn't matter which rule he was talking about. – tarit goswami Feb 08 '19 at 07:46
  • See related https://math.stackexchange.com/a/1782225/72031 – Paramanand Singh Feb 08 '19 at 18:16

3 Answers3

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As stated in the comments we can use the fact that $$x^n=e^{n\ln{x}}$$ For all $x \in \mathbb{C}$ except $0$, $n \in \mathbb{C}$. Then, by using the chain rule, we have $$\frac{d}{dx}\Big(e^{f(x)}\Big)=f'(x)\cdot e^{f(x)}$$

So, $$\frac{d}{dx}\Big(x^n\Big)=\frac{d}{dx}\Big(e^{n\ln{x}}\Big)=\frac{n}{x}\cdot e^{n\ln{x}}=\frac{n}{x} \cdot x^n = n \cdot x^{n-1}$$

YiFan Tey
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Peter Foreman
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For $\boldsymbol{n\ge1}$

Bernoulli's Inequality, which is proven for integer exponents in this answer and extended to rational exponents in this answer using induction, says that for $n\ge1$, $$ 1+nx\le(1+x)^n\tag1 $$ From $(1)$, we get $$ 1+x\le\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x\tag2 $$ Therefore, $$ x^n\left(1+\frac{nh}x\right)\le\overbrace{x^n\left(1+\frac hx\right)^n}^{(x+h)^n}\le x^ne^{nh/x}\tag3 $$ where the left inequality is $(1)$ and the right inequality is the $n^\text{th}$ power of $(2)$.

Subtracting $x^n$ and dividing by $h$ gives $$ x^n\frac nx\stackrel{h\gtrless0}\lesseqgtr\frac{(x+h)^n-x^n}{h}\stackrel{h\gtrless0}\lesseqgtr x^n\frac nx\frac{e^{nh/x}-1}{nh/x}\tag4 $$ Applying $(9)$ gives $$ x^n\frac nx\stackrel{h\gtrless0}\lesseqgtr\frac{(x+h)^n-x^n}{h}\stackrel{h\gtrless0}\lesseqgtr x^n\frac nx\frac1{1-nh/x}\tag5 $$ Then the Squeeze Theorem yields $$ \bbox[5px,border:2px solid #C0A000]{\frac{\mathrm{d}}{\mathrm{d}x}x^n=nx^{n-1}}\tag6 $$

Extending to $\boldsymbol{n\lt1}$

For smaller $n$, we can use the product rule and induction. That is, suppose we know that $(6)$ holds for some $n$, then $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^{n-1} &=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac1xx^n\right)\\ &=\frac1xnx^{n-1}-\frac1{x^2}x^n\\ &=(n-1)x^{n-2}\tag7 \end{align} $$ thus, $(6)$ holds for $n-1$.

Bounds on $\boldsymbol{\frac{e^x-1}x}$

Taking $(2)$, substituting $x\mapsto-x$, and taking reciprocals yields that for $x\lt1$, $$ e^x\le\frac1{1-x}\tag8 $$ Combining $(2)$ and $(8)$, subtracting $1$ and dividing by $x$ gives $$ 1\stackrel{x\gtrless0}\lesseqgtr\frac{e^x-1}x\stackrel{x\gtrless0}\lesseqgtr\frac{1}{1-x}\tag9 $$

robjohn
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  • Pretty neat. Very Informative. Thank You. By the way, is it possible to extend Bernoulli's inequality to irrational numbers as well ? –  Feb 08 '19 at 14:28
  • In the same way as always: by continuity. – robjohn Feb 08 '19 at 14:30
  • I have come across this term ("continuity argument") a lot while studying Inequalities, for example when generalizing the weighted AM-GM ineqality for all real weights. But I do not understand (at all) what is meant by this term. I will be very grateful if you can provide some links or references to explore. –  Feb 08 '19 at 14:34
  • We define $x^n$ for integer values of $n$ by repeated multiplication and division. Then we define $x^{n/m}$ for rational $n/m$ by defining $y=x^{n/m}$ where $y^m=x^n$. We then define $x^\alpha$ for real $\alpha$ by continuity; that is, $x^\alpha=\lim\limits_{m/n\to\alpha}x^{m/n}$. – robjohn Feb 08 '19 at 15:29
  • That is, we define $x^\alpha$ for $\alpha\in\mathbb{R}$ to be the continuous extension of $x^{n/m}$ for $n/m\in\mathbb{Q}$. – robjohn Feb 08 '19 at 15:49
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The next step is to prove the result for exponents of the form $1/n \text{ with } n \in \Bbb N \text{ and } n \gt 0$. You can do this via implicit differentiation: If $y = x^{1/n}, \text{ then } y^n = x$. Then use the quotient rule to get the result for negative exponents and the chain rule and the two previous results (using $x^{p/q} = {(x^{1/q})}^p$) to obtain the result for all rational exponents.

Obtaining the result for irrational exponents first requires you to define exponentiation for irrational exponents. You would get a correct definition by taking limits of rational exponents, but the definition that's much easier to work with is $x^n=e^{n \cdot\ln x}$, which you can differentiate using the chain rule and the fact (derived from the definition of $e^x$) that $d(e^x)/dx=e^x$.

And by the way, good job realizing that using a Taylor series would be circular reasoning.

Robert Shore
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