# Prove that the derivative of $x^w$ is $w x^{w-1}$ for real $w$

Can anyone give a proof of the derivative of this type of function? Specifically showing, $\frac{d(x^w)}{dx} = wx^{w-1}$ for a real $w$?

I tried to use the Taylor series expansion for $(x+dx)^w$ and got the correct result. However, the proof of the Taylor series requires knowledge of the derivative of these functions. So this is essentially circular reasoning. I know that the same series is also given by the binomial expansion, but that's not entirely satisfactory either, because where's the proof that the binomial expansion works for all reals (isn't it only apparent for integers)? So far all of the arguments I've come across involve circular reasoning.

I was thinking of showing that the binomial expansion is true for all reals using some form of proof by induction e.g. something like this. http://www.math.ucsd.edu/~benchow/BinomialTheorem.pdf I'm really not sure.

## 7 Answers

Write $x^n=e^{n\log x}$ (no restriction on $n$ being an integer) and use the chain rule to show that $$\frac{d(e^{n\log x})}{dx}=e^{n\log x}\times\frac{n}{x}=nx^{n-1}.$$

• The only downside to this approach is it assumes the derivative exists. – zz20s Jul 11 '16 at 16:15
• This is the only answer that addresses what the original question seems to be asking--show that the power rule holds for exponents which are not positive integers. – Jakob Hansen Jul 11 '16 at 16:19
• thanks, I don't think the proof for the derivative of e^x requires the derivative of p series. It's simply $\frac{d(e^x)}{dx} = e^x \cdot{ f'(0)} = e^x\cdot{c}$, selecting e such that c = 1. That's fairly rigorous, I think.. – Dis-integrating Jul 11 '16 at 16:25
• If you define $\log(x) = \int_1^x \frac{1}{t}\,dt$ and $\exp$ as the inverse of $\log$, all you need to find the derivative of $\exp$ is the inverse function theorem. And, I suppose, the fundamental theorem of calculus. – Jakob Hansen Jul 11 '16 at 16:32
• @JakobHansen Yet, then one needs to show that $\log(x^n)=n\log(x)$ for real values of $n$. If $n\mathbb{Q}$, this is straightforward using the integral definition of the logarithm function. But to extend to the reals, one needs to invoke a density argument, which is not difficult, but does require some more work. – Mark Viola Feb 9 at 21:08

you want to calculate $\lim\limits _{h\to 0}\frac{(x+h)^n-x^n}{h}$

use the binomial theorem:

$$\lim\limits _{h\to 0}\frac{(x+h)^n-x^n}{h}=\lim\limits _{h\to 0}\frac{\sum_{k=0}^n\binom{n}{k}x^kh^{n-k}-x^n}{h}=$$

$$\lim\limits _{h\to 0} nx^{n-1}+h(\binom{n}{2}x^{n-2}+\dots+\binom{n}{n}h^{n-1})$$

The part in the right clearly goes to zero.

• This for me is the best proof because the limit in the infinitesimal change in $x^n$ per infinitesimal change in $x$ is the definition of differentiation; the limit of difference quotients – samerivertwice Jul 11 '16 at 16:14
• then what is the proof of the binomial theorem for reals? – Dis-integrating Jul 11 '16 at 16:27
• it's purely algebraic, by induction, it works for any commutative ring. – Jorge Fernández Hidalgo Jul 11 '16 at 16:28
• and if you only want it for $(1+x)^n$ it works for any ring. – Jorge Fernández Hidalgo Jul 11 '16 at 16:29
• The problem is that OP wants $n$ to be real. – quid Jul 11 '16 at 16:30

$$y = x^n$$ $$\ln(y) = n\ln(x)$$ $$\frac{y'}{y} = \frac{n}{x}$$ $$y' = nx^{n - 1}$$ The derivatives of logarithms are defined by definition.

• "defined by definition"? This is also a (the second) satisfactory proof but essentially equivalent to the accepted answer. – Dis-integrating Jul 11 '16 at 17:35
• who posted first – Dis-integrating Jul 11 '16 at 17:35
• "Defined by definition" being that the logarithm can be defined as the antiderivative of 1/x. The other answerer posted first. And you're right, the answers are equivalent. I just used safer derivatives. – Kaynex Jul 12 '16 at 5:51
• He deserves the "correct answer". I simply posted this as this is the method I would personally use. – Kaynex Jul 12 '16 at 6:11

I wrote an answer to a question that was closed as a duplicate of this one. I thought I would add a different answer to this question.

Integer Case

For integer $$n\ge0$$, the Binomial Theorem says $$(x+h)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}h^k\tag1$$ So $$\frac{(x+h)^n-x^n}h=\sum_{k=1}^n\binom{n}{k}x^{n-k}h^{k-1}\tag2$$ Therefore, taking the limit, we get \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^n &=\binom{n}{1}x^{n-1}\\[3pt] &=nx^{n-1}\tag3 \end{align}

Inverting

Suppose that $$x=y^m\tag4$$ Taking the derivative of $$(4)$$ using $$(3)$$ and substituting $$y=x^{\frac1m}$$ yields $$1=my^{m-1}\frac{\mathrm{d}y}{\mathrm{d}x}=mx^{1-\frac1m}\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac1m}\tag5$$ Therefore, $$\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac1m}=\frac1mx^{\frac1m-1}\tag6$$

Rational Case

Applying the Chain Rule with $$(3)$$ and $$(6)$$ gives \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^{\frac nm} &=n\left(x^{\frac1m}\right)^{n-1}\frac1mx^{\frac1m-1}\\[3pt] &=\frac nmx^{\frac nm-1}\tag7 \end{align}

Real Case

When a sequence of functions converges pointwise and their derivatives converge uniformly, the derivative of the limit equals the limit of the derivatives (see this question). Therefore, the full case for non-negative exponents follows by continuity.

Negative Case

Applying the Chain Rule with $$(7)$$ and $$\frac{\mathrm{d}}{\mathrm{d}x}\frac1x=-\frac1{x^2}$$, we get \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^{-s} &=-\frac1{(x^s)^2}sx^{s-1}\\ &=-sx^{-s-1}\tag8 \end{align}

• Don’t you need derivatives to prove binomial theorem for negative integers? To fix this, you could differentiate x^-1 directly using limits and extend it to negatives using the product rule. But since it’s the same pattern for all reals, I feel that the best proof does it all in one clean swoop, so far the only instance of that has been to derive it from exponential derivatives – Dis-integrating Feb 10 at 23:19
• It can be done without derivatives, but I have added something for negative exponents. – robjohn Feb 10 at 23:31
• This argument is mathematically more interesting imo – Dis-integrating Feb 10 at 23:38

Hint
Use induction along with the product rule.

For real $n$,

$$\lim_{h\to0}\frac{(x+h)^n-x^n}n=\lim_{h\to0}\frac{\left(1+\frac hx\right)^n-1}hx^n=\lim_{h\to0}\frac{\left(1+\frac hx\right)^n-1}{\frac hx}x^{n-1}=\phi(n)x^{n-1},$$ as the last limit cannot depend on $x$.

Then from

$$(x^{n+m})'=(x^nx^m)'$$ you can deduce the linearity

$$\phi(n+m)=\phi(n)+\phi(m).$$

With the obvious $\phi(1)=1$, this should be enough to prove that $\phi$ is the identity.

• This is a really nice proof, I was convinced that the answer lied in the fact that $x^{n+m} = x^nx^m$ for all n and m, but couldn't figure out a proof. All the correct answers have used it so far so that's probably what it is. Thanks.. – Dis-integrating Jul 13 '16 at 21:08
• This only proves that if the function is differentiable and if $x^{n+m}=x^nx^m$, then $\phi(n)$ is defined, $\phi(n+m)=\phi(n)+\phi(m)$ and $\phi(1)=1$. But you have to do more work from deducing from $\phi(m+n)=\phi(m)+\phi(n)$ and $\phi(1)=1$ that $\phi(n)=n$: you're basically assuming $\phi$ is continuous. – egreg Jul 17 '16 at 9:06
• @egreg: right, this is not a complete proof. As usual, you don't know what are the preliminaries of the question. – Yves Daoust Jul 21 '16 at 17:28

I saw a lot of talk in comments (and comments moved to chat) about the Taylor series. I also indicated an answer in my comments which avoids Taylor's series and instead proves derivative of $x^{n}$ using basic algebra.

From OP's post it is clear that an answer based on binomial theorem which holds for general real index is desired. View point of OP is correct, but requires some more effort to establish the general binomial theorem and my post below does exactly that.

In what follows we assume that $x > 0, n \in \mathbb{R}$. The questions asks us to prove $$\frac{d}{dx}x^{n} = nx^{n - 1}\tag{1}$$ Note that when $n$ is irrational the symbol $x^{n}$ can not be handled by algebra and hence it is useless to expect a proof which is based on algebra alone.

However when $n$ is rational then $x^{n}$ is an algebraic function and it makes sense to have an (almost) algebraic proof. So we first deal with the simpler case when $n$ is rational. One of the approaches is to use definition of derivative and focus on the ratio $$\frac{(x + h)^{n} - x^{n}}{h}$$ and here we can write $$(x + h)^{n} = x^{n}(1 + h/x)^{n}$$ and then apply general binomial theorem to expand $(1 + h/x)^{n}$ as an infinite series. There are two viewpoints regarding this approach:

• This approach appears circular because expansion of $(1 + h/x)^{n}$ when $n$ is not a positive integer essentially requires Taylor series and involves the derivative of $x^{n}$. This point of view is almost correct.
• It is possible to establish the general binomial expansion of $(1 + h/x)^{n}$ even when $n$ is not a positive integer without the use of derivatives. This is the point of view of this answer.

Let $n \in \mathbb{R}, |x| < 1$ so that the series $$f(x, n) = 1 + nx + \frac{n(n - 1)}{2!}x^{2} + \frac{n(n - 1)(n - 2)}{3!}x^{3} + \cdots\tag{2}$$ is absolutely convergent and $f(x, n)$ is well defined.

Let $p, q$ be positive integers then we know from the binomial theorem for positive integral index that $$f(x, p) = (1 + x)^{p}, f(x, q) = (1 + x)^{q}, f(x, p + q) = (1 + x)^{p + q}$$ and therefore $$f(x, p)f(x, q) = f(x, p + q)$$ and considering the coefficients of $x^{r}$ on both sides we see that $$\binom{p + q}{r} = \binom{p}{0}\binom{q}{r} + \cdots + \binom{p}{i}\binom{q}{r - i} + \cdots + \binom{p}{r}\binom{q}{0}\tag{3}$$ where we have by definition $$\binom{a}{0} = 1, \binom{a}{r} = \frac{a(a - 1)(a - 2)\cdots (a - r + 1)}{r!}\tag{4}$$ for all real $a$ and positive integer $r$ so that the general binomial coefficient is actually a polynomial in $a$.

We can now see that the identity $(3)$ is an identity which involves polynomials in two variables $p, q$ and it holds for any infinity of values of $p, q$ (it holds for all positive integers $p, q$) and hence it holds identically. Therefore the identity $(3)$ is true for all variables $p,q$. Now by multiplication of infinite series we see that the following identity holds for all real variables $p, q$ and $|x| < 1$: $$f(x, p)f(x, q) = f(x, p + q)\tag{5}$$ and thus $f(x, p)$ behaves like an exponential function as far as parameter $p$ is concerned. It follows by the use of the above functional equation that $$f(x, n) = \{f(x, 1)\}^{n}$$ if $n$ is rational. Hence we have $$(1 + x)^{n} = 1 + nx + \frac{n(n - 1)}{2!}x^{2} + \frac{n(n - 1)(n - 2)}{3!}x^{3} + \cdots\tag{6}$$ for all rational values of $n$. This means that the general binomial theorem for rational index can be proved without any use of derivatives and using this we can establish the derivative of $x^{n}$ for rational $n$.

What happens when $n$ is irrational? The first problem is to define $x^{n}$ for irrational $n$ and there are many approaches and the simplest one is define it as $x^{n} = \exp(n\log x)$. With this definition it is easy to prove derivative formula using derivatives of exponential and logarithmic functions.

Another approach is to define $x^{n}$ for irrational $n$ via continuity. Thus if $n_{k}$ is a sequence of rationals tending to irrational $n$ as $k \to \infty$ then we define $x^{n}$ to the limit of $x^{n_{k}}$ as $k \to \infty$. If we adopt this definition then we can prove with some effort that the function $f(x, n)$ is a continuous function of $n$ for all real $n$ and fixed $x$ with $|x| < 1$. And hence by continuity the general binomial theorem holds for all real index $n$ and our problem of calculating derivative of $x^{n}$ is handled in usual manner indicated above.

• no need for the binomial coefs and all those details. It is obvious $(x+h)^n$ is a polynomial in $h$ and that the two first terms are $x^n+nh x^{n-1}$ – reuns Jul 17 '16 at 7:03
• @user1952009: A typo in your comment. It should be $$(x + h)^{n} = x^{n} + nhx^{n - 1} + O(h^{2})$$ But how do you prove it without using any sort of derivatives. The only way out is to use some algebra of inequalities (see the linked answer) or to use binomial theorem for general index. It also appears from your comment that you think $n$ to be a positive integer. My answer deals with the case when $n$ is not a positive integer. – Paramanand Singh Jul 17 '16 at 7:05
• @user1952009: I don't think any of my details are unnecessary. remove any stuff and the answer becomes wrong. Let me know which step you think is unnecessary. – Paramanand Singh Jul 17 '16 at 7:09
• yes your post messed me up with all those unnecessary details. and $n = 1/q$ is handled by the derivative of the inverse function $(f^{-1}(x))' = \frac{1}{f^{-1}(f'(x))}$, and $a$ real as the limiting case of $p/q$ – reuns Jul 17 '16 at 7:09
• @user1952009: when you use inverse function stuff you are already dealing with derivative laws. The answer I gave uses just the bare minimum definition of derivative. The approach using derivative law is short and smart but not exactly as desired by OP. His approach to use the definition of derivative and not derivative laws. – Paramanand Singh Jul 17 '16 at 7:12