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

Question

Can anyone give a proof of the derivative of this type of function? Specifically showing that

$\dfrac{d(x^w)}{dx} = wx^{w-1}$ 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.

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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.

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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 $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}x^n &=\binom{n}{1}x^{n-1}\\[3pt] &=nx^{n-1}\tag3 \end{align} $$

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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, $$ \frac{\mathrm{d}}{\mathrm{d}x}x^{\frac1m}=\frac1mx^{\frac1m-1}\tag6 $$

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Rational Case

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

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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.

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Negative Case

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

Answer 5

We can use a method called logarithmic differentiation: \begin{align} y&=x^n\\ \log y&=\log x^n\\ \log y&=n\log x\\ \frac{d}{dx}\left(\log y\right)&=\frac{d}{dx}\left(n\log x\right)\\ \frac{y'}{y}&=\frac{n}{x}\\ y'&=\frac{n}{x}(y)\\ y'&=\frac{n}{x}(x^n)\\ y'&=nx^{n-1} \end{align} This uses the fact that $$\frac{d}{dx}\left(\log x\right)=\frac1x.$$

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To see why this works for the part of the domain $x<0$ (since we are working with logarithms), we can consider that $$x=(-1)(-x)\implies x^n=(-1)^n(-x)^n$$ and we arrive at the same result, using the definition of the complex logarithm: \begin{align} y&=(-1)^n(-x)^n\\ \log y&=\log(-1)^n+\log(-x)^n\\ \log y&=\log(-1)^n+\log(|-x|)^n+i\pi\\ \end{align} Here, on the right side of the equation, the first and last terms are constants, which yield a derivative of $0$.

Answer 6

Hint
Use induction along with the product rule.

Answer 7

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.

Answer 8

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.

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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.

Answer 9

Here is one more suggestion without use of logarithms.

You know $\frac{d}{dx} x^n=nx^{n-1}$ for all integers, including when $n$ is negative.

Now let us get one step further extending this to rational numbers by asking ourselves what is $\frac{d}{dx} x^{\frac{p}{q}}$ where $p$ and $q\neq 0$ are integer numbers.

Let $y=x^{\frac{p}{q}}$ and then take power to $q$ on both sides so that you get $y^q=x^p$. Now take derivate with respect to $x$ on both sides to get

$$qy^{q-1}\frac{dy}{dx}=px^{p-1} \Leftrightarrow \frac{dy}{dx}=\frac{px^{p-1}}{qy^{q-1}}=\frac{p}{q}x^{p-1}y^{1-q}=\frac{p}{q}x^{p-1+\frac{p}{q}-p}=\frac{p}{q}x^{\frac{p}{q}-1}.$$

To extend to the final case $y=x^a$ for irrational $a$ you need to take limits. For simplicity, let $x$ be positive and to show $\frac{dy}{dx}=ax^{a-1}$ you take the limit from both sides of the rational case such that

$$\frac{dy}{dx}x^{a_{-}}<\frac{dy}{dx}x^{a}<\frac{dy}{dx}x^{a_{+}}$$

where rational number $a_{-}=\max\{\frac{p}{q}:\frac{p}{q}<a, p,q\in\mathbb N\}$ and rantional number $a_{+}=\min\{\frac{p}{q}:\frac{p}{q}>a, p,q\in\mathbb N\}$. Since limits $x^{a_{-}}=x^{a_{+}}$ we have finished our proof.

Answer 10

Some feeling of the existing awkwardness prompts me to write own version, in hope it, also will be useful. I will start the same way as Yves Daoust, but continue in a different way

$$\lim_{h\to0}\frac{(x+h)^n-x^n}n=x^{n-1}\lim_{h\to0}\frac{\left(1+\frac hx\right)^n-1}{\frac hx} = x^{n-1}\lim_{\alpha \to0}\frac{\left(1+\alpha\right)^n-1}{\alpha}$$ Now let's take $\beta = (1+\alpha)^n-1$, then we have, by continuity, when $\alpha \to 0$, then $ \beta \to 0$ and $$\frac{\left(1+\alpha\right)^n-1}{\alpha} = \frac{\beta}{\alpha} = \frac{\beta}{\ln (1+ \beta)} \cdot n \cdot \frac{\ln (1+\alpha)}{\alpha}$$ now, knowing by logarithm continuity, that $\frac{\beta}{\ln (1+ \beta)}$ and $\frac{\ln (1+\alpha)}{\alpha}$ tends to $1$, we obtained desired.