# How can I prove $\frac {d}{dx} {x^n} = n x^{n-1}$ for $n \in \Bbb R$ without circular reasoning? [duplicate]

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 ?

• How about this formula: $x^n=\exp(n\ln x)$? – Angina Seng Feb 8 '19 at 7:34
• @LordSharktheUnknown But this will hold only for $x > 0$. – user399078 Feb 8 '19 at 7:38
• @Nirbhay: if you are interested in non-integer $n$, then there are difficulties in considering $x^n$ for $x\lt0$. – robjohn Feb 8 '19 at 7:40
• @robjohn Yeah got it. Thanks. – user399078 Feb 8 '19 at 7:41
• @taritgoswami I know all the "rules" and stuff. But will the product rule be useful ? – user399078 Feb 8 '19 at 7:42

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

• How to prove that $\frac{d}{dx}\Big(e^{f(x)}\Big)=f'(x)\cdot e^{f(x)}$? – Three Sided Coin Feb 8 '19 at 8:33
• The product rule states that $\frac{d}{dx} f(g(x)) = f'(g(x))g'(x)$ and the derivative of $e^x$ is $e^x$ – Peter Foreman Feb 8 '19 at 8:43
• @PeterForeman: that looks more like the chain rule. – robjohn Feb 8 '19 at 10:37
• Sorry, that's what I meant. – Peter Foreman Feb 8 '19 at 13:40
• That solved my problem ! Thank You. – user399078 Feb 8 '19 at 14:29

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 $$1\stackrel{x\gtrless0}\lesseqgtr\frac{e^x-1}x\stackrel{x\gtrless0}\lesseqgtr\frac{1}{1-x}\tag9$$

• Pretty neat. Very Informative. Thank You. By the way, is it possible to extend Bernoulli's inequality to irrational numbers as well ? – user399078 Feb 8 '19 at 14:28
• In the same way as always: by continuity. – robjohn Feb 8 '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. – user399078 Feb 8 '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 8 '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 8 '19 at 15:49

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.