Elementary proof of Power Rule For differentiation $f'(x)=rx^{r-1}$ for $f(x)=x^r$

Question

I could not find any elementary proof of the Power Rule for differentiation:

Given $x,r \in R$ ,$x>0$ and a function $f(x)=x^r$, then its derivative is $f'(x)=rx^{r-1}$

Defintion:Let a sequence of rational numbers {$r_n$} tend to $r$ then $x^r=\lim_{r_n \to r}x^{r_n}$

All the proofs I had seen utilized the derivative of $logx$ but I dont think this is elementary because one has to show that for $e=\text{Euler's number}$ , $t \in R$and $n \in N$ $\lim_{n \to \infty}(1+\frac{1}{n})^n=\lim_{t \to 0}(1+t)^\frac{1}{t}=e$ which can be proved using the Power Rule

Ofcourse the proof for the Power Rule is straight forward if one is familiar with the result of Power Rule if $r$ is a rational number and theorem regarding derivative of a sequence of derivatives that are uniformly convergent.

Answer 1

For real exponents in general, I don't think you can avoid the exponential function and the logarithm because the definition of $x^r$ tends to be $\exp(r\log x)$.

However, for integer exponents there are a number of simple proofs. Here is a lesser-known one. Let $f:\mathbb{R}\mapsto\mathbb{R}$ be the function defined by $f(x)=x^n$ for all $x$, with $n \in \mathbb{N}$. We find that \begin{align} f'(a) &= \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \\[6pt] &= \lim_{x \to a}\frac{x^n-a^n}{x-a} \\[6pt] &= \lim_{x \to a}\frac{(x-a)(x^{n-1}+x^{n-2}a+x^{n-3}a^2+\ldots+xa^{n-2}+a^{n-1})}{x-a} \\[6pt] &= \lim_{x \to a}x^{n-1}+x^{n-2}a+x^{n-3}a^2+\ldots+xa^{n-2}+a^{n-1} \\[6pt] &= a^{n-1}+a^{n-2}a+a^{n-3}a^2+\ldots+aa^{n-2}+a^{n-1}\\[6pt] &= \underbrace{a^{n-1}+a^{n-1}+a^{n-1}+\ldots+a^{n-1}+a^{n-1}}_{\text{$n$ terms}}\\[6pt] &= na^{n-1}\blacksquare \end{align} This proof can be extended to negative integer exponents by writing $x^{-n}$ as $1/x^n$ and using the quotient rule.

Answer 2

For $r\in\mathbb{R}$, to prove that $f(x)=x^{r}\implies f'(x)=rx^{r-1}$ We need two lemmas :

Lemma $(1)$ : Consider the function $E(x)=\displaystyle\sum_{n=0}^{+\infty}\frac{x^{n}}{n!}$. We claim that $E'(x)=E(x)$.

Proof : Let $x\in\mathbb{R}$ with $h\neq0$, we define $\tau(h)=\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{E(x+h)-E(x)}{h}$. Since $E(x+y)=E(x)E(y)$, then we have that : $$ \tau(h)=\displaystyle\frac{f(x+h)-f(x)}{h}=\frac{E(x+h)-E(x)}{h}=E(x)\cdot\frac{E(h)-1}{h} $$ Thus we need to prove that $\displaystyle\lim_{h\to 0}\frac{E(h)-1}{h}=1$. Note that : \begin{align*} \frac{E(h)-1}{h}&=\frac{1}{h}\left(\left(1+h+\frac{h^{2}}{2!}+\cdots\right)-1\right) \\ &=1+\frac{h}{2!}+\frac{h^{2}}{3!}+\frac{h^{3}}{4!}+\cdots \\ \implies\left|\frac{E(h)-1}{h}-1\right|&=\left|\frac{h}{2!}+\frac{h^{2}}{3!}+\frac{h^{3}}{4!}+\cdots\right| \\ &\leq\frac{|h|}{2!}+\frac{|h|^{2}}{3!}+\frac{|h|^{3}}{4!}+\cdots \\ &=\underbrace{|h|}_{\to0}\underbrace{\left(\frac{1}{2!}+\frac{|h|}{3!}+\frac{|h|^{2}}{4!}+\cdots\right)}_{\text{bounded}} \end{align*} Since $\displaystyle\lim_{h\to 0}|h|=0$, and if we can show that the other part is bounded, then the overall limit is $0$. Let $0<|h|<1$, we note that the chosen interval of $h$ is due to our interest of $h$ being close to $0$. We have that : \begin{align*} &\left|\frac{E(h)-1}{h}-1\right|=|h|\left(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\right)=|h|\underbrace{\sum_{n=2}^{+\infty}\frac{1}{n!}}_{\text{converges}} \\ \implies\lim_{h\to 0}&\left|\frac{E(h)-1}{h}-1\right|=\lim_{h\to 0}|h|\sum_{n=2}^{+\infty}\frac{1}{n!}=0 \end{align*} Therefore, by sandwich theorem : $$ \lim_{h\to 0}\left|\frac{E(h)-1}{h}-1\right|=0\implies\lim{h\to 0}\frac{E(h)-1}{h}=1 $$ Hence $\displaystyle\lim_{h\to 0}\tau(h)=\lim_{h\to 0}E(x)\cdot\frac{E(h)-1}{h}=E(x)$, $\forall x\in\mathbb{R}$ with $E'(x)=E(x)$.

Lemma $(2)$ : Let $g(x)=\ln(x)$, we claim that $\displaystyle g'(x)=\frac{1}{x}$, $\forall x>0$.

Proof : Let $x_{0}\in(0,+\infty)$ with $x_{0}\neq x$, let $\displaystyle\phi(x):=\frac{\ln(x)-\ln(x_{0})}{x-x_{0}}$. Moreover, there exists $t_{x}\in\mathbb{R}$ with $e^{t_{x}}=x$ and there exists $t_{0}\in\mathbb{R}$ with $e^{t_{0}}=x_{0}$. Thus, we have that :$$\displaystyle\phi(x)=\frac{t_{x}-t_{0}}{e^{t_{x}}-e^{t_{0}}}=\displaystyle\frac{1}{\displaystyle\frac{e^{t_{x}}-e^{t_{0}}}{t_{x}-t_{0}}}$$ Notice that $\ln(x)$ is continuous and that $\ln(x)=t_{x}$ and $\ln(x_{0})=t_{0}$. Thus, we have that : \begin{align*} &\lim_{x\to x_{0}}\ln(x)=\ln(x_{0})=t_{0} \\ \implies&\lim_{x\to x_{0}}\phi(x)=\lim_{x\to x_{0}}\frac{\ln(x)-\ln(x_{0})}{x-x_{0}}\\ &\;\qquad\qquad=\frac{1}{(e^{t})'|_{t=t_{0}}}=\frac{1}{e^{t_{0}}}=\frac{1}{x_{0}} \end{align*} Therefore, $g$ is differentiable at $x_{0}$ and $g'(x_{0})=\displaystyle\frac{1}{x_{0}}$, $\forall x_{0}\in(0,+\infty)$.

Conclusion : We have that $f(x)=x^{r}=e^{\ln(x^{r})}=e^{r\ln(x)}$. We note that $f$ is the composition of $e^{x}$ and $r\ln(x)$ that are both differentiable. Thus : $$ f'(x)=(r\ln(x))'(e^{x})'|_{x=r\ln(x)} =\frac{r}{x}e^{r\ln(x)}=\frac{r}{x}x^{r}=rx^{r-1} $$

Answer 3

For positive integer $r$, we can use induction.

True for $r=1$: $(x^1)' = x' = 1 = 1x^0$.

If true for $r$, $(x^{r+1})' = (x\ x^r)' = x'x^r+x(x^r)' =x^r+x(rx^{r-1}) =x^r+rx^r = (r+1)x^r $.

Answer 4

You're not going to get an elementary proof for real $r$ in general. But you can go a long way by elementary means.

natural $r$:

Use induction starting at $f(x)=x$ and then use the product rule for the inductive step. For $r=0$ the result is trivial.

integer $r$:

Use that $x^{-r}=(x^r)^{-1}$, then apply the chain rule. This way you only need to prove $\frac{\mathrm dx^{-1}}{\mathrm dx}=-x^{-2}$. Which goes like this:

$$\begin{align}\frac{(x+h)^{-1}-x^{-1}}{h}&=\frac{\frac{x}{x(x+h)}-\frac{x+h}{x(x+h)}}{h}\\ &=\frac{-\frac{h}{x(x+h)}}{h}\\ &=-\frac{1}{x(x+h)},\end{align}$$

which goes to $-x^{-2}$ as $h\to0$.

rational $r$:

Use that $x^{\frac pq}=(x^p)^{\frac1q}$ and the chain rule, so you only need to prove the fact for $r=\frac1q,~q\in\mathbb N_{\geq2}$. Which follows from the rule for the derivative of the inverse function, since $x^{\frac1q}$ is the inverse of $x^q$. Exception: at $x=0$ the inverse function rule doesn't work since we would divide by $0$. But for $x=0$, the derivative is again elementary, since $\frac{h^q}{h}=h^{q-1}$ goes to $0$ as $h\to0$ for all $q>1$.

irrational $r$:

This is where elementary means won't help us anymore. But it can be done without logarithms. Let's say $r_n\to r$, where $r_n$ are all rational. We want to calculate the limit

$$\lim_{h\to0}\frac{\lim\limits_{n\to\infty}(x+h)^{r_n}+\lim\limits_{n\to\infty}x^{r_n}}{h}.$$

Using the limit laws we can manipulate this to get

$$\lim_{h\to0}\lim_{n\to\infty}\frac{(x+h)^{r_n}-x^{r_n}}{h}.$$

If we could exchange the two limits, we would get that this is equal to

$$\lim_{n\to\infty}\lim_{h\to0}\frac{(x+h)^{r_n}-x^{r_n}}{h}=\lim_{n\to\infty}\frac{\mathrm dx^{r_n}}{\mathrm dx}=\lim_{n\to\infty}r_n x^{r_n-1}=r x^{r-1},$$

where the last step uses elementary limit laws and your definition of irrational powers. But we need to be able to exchange the limits. There is a theorem which says that if both $\lim_{h\to0} a_n(h)$ and $\lim_{n\to\infty}a_n(h)$ exist and the latter limit is uniform if we consider $a_n$ as a sequence of functions of $h$, then we are allowed to exchange the limits. But I'd classify this theorem as similarly non-elementary as logarithms. Anyway, we now need to prove that

$$\frac{(x+h)^{r_n}-x^{r_n}}{h}$$

converges uniformly as a sequence of functions of $h$. I'm having a hard time doing so right now, but I'm pretty optimistic that it can be proved. You might want to try it as an exercise.

Answer 5

I think this is elementary as it requires:

1-Definition of continuity and derivative

2-Binomial Theorem ( which I think is quite elementary)

3-Geometric series

4-Series convergence test by comparison

5-limit of Uniformly Convergent series is continuous

It suffice to prove $\frac{d}{dx}\log{x}=\frac{1}{x}$ using elementary results

$x > 0$ , $n \in N$

define $e=\lim_{n \to \infty}(1+\frac{1}{n})^n$ which exists according to binomial theorem and comparison with geometric series

define $g(x)=\lim_{n \to \infty}(1+\frac{x}{n})^n$ which exists as shown below

$h(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$ which exists by comparison with geometric series and is uniformly convergent

using binomial theorem $g(x)=h(x)$

because $x>0$, $g(x)$ is increasing on $[0,\infty)$

$k,m,j \in N$

$ y \in Q$

$y=\frac{k}{m}$

$g(y)=\lim_{n \to \infty}((1+\frac{y}{n})^{\frac{n}{y}})^y$

set $n=kj$

$g(y)=\lim_{j \to \infty}((1+\frac{1}{mj})^{mj})^y=e^y$

because $g(x)$ is increasing for every $x>0$ we can find two rational numbers $p,q$ and $q>p$ such that $g(p)\le g(x) \le g(q)$

since $h(x)$ is continuous the limit $\lim_{r_n \to r}e^{r_n}=\lim_{r_n \to r}h(r_n)=h(r)$ , where $\{r_n\}$ is a sequence of rational numbers converging to $r$

So it makes sense to define $e^r=\lim_{r_n \to r}e^{r_n}$

since we can let $ p \to q$ and $e^x$ is continuous ,this implies $e^x=h(x)$

$\frac{dh}{dx}=\lim_{t \to 0} e^x\frac{e^t-1}{t}=\lim_{t \to 0} e^x\frac{\sum_{n=1}^\infty \frac{t^n}{n!}}{t} = e^x$

$x=e^{\log{x}}$

taking derivative of both sides and using chain rule

$1=e^{\log{x}}\frac{d}{dx}\log{x}$

$\frac{d}{dx}\log{x}=\frac{1}{e^{\log{x}}}=\frac{1}{x}$

Answer 6

Possibly the most "elementary" way to do it is to define logarithms first, and only later define general exponentials (and not to do it the way you do, except to show the new definition agrees with the standard one for integers and rationals).

For $x\gt 0$, define the function $\log(x)$ be $$\log(x) = \int_1^x \frac{1}{t}\,dt.$$

The following properties are now straightforward:

1. $\log(x)$ is strictly increasing and differentiable.
2. $\frac{d}{dx}\log(x) = \frac{d}{dx}\int_1^x\frac{1}{t}\,dt = \frac{1}{x}$.
3. $\lim_{x\to0^+}\log(x) = -\infty$ and $\lim_{x\to \infty}\log(x)=\infty$.
4. $\log(1)=0$.

Then we also have, by letting $u=\frac{1}{t}$, that $$\begin{align*} \log\left(\frac{1}{x}\right) &= \int_1^{1/x}\frac{1}{t}\,dt &= \int_1^x u\left(-\frac{u}{u^2}\right)\,du\\ &= -\int_1^x\frac{1}{u}\,du\\ &= -\log(x). \end{align*}$$

Also: fix $a\gt 0$. Then letting $au=t$, we have $$\begin{align*} \log(ax) &= \int_1^{ax}\frac{1}{t}\,dt\\ &= \int_{1/a}^x\frac{1}{au}a\,du\\ &= \int_{1/a}^x\frac{1}{u}\,du = -\int_1^{1/a}\frac{1}{u}\,du + \int_1^x\frac{1}{u}\,du\\ &= -\log\left(\frac{1}{a}\right) + \log(x)\\ &= \log(a)+\log(x). \end{align*}$$ So now we also have the following further properties: for $a,b\gt 0$,

5. $\log(ab)=\log(a)+\log(b)$.
6. $\log(a/b)=\log(a)-\log(b)$.

By induction, it follows that for positive integers $n\gt 0$, $\log(a^n) = n\log(a)$; and that $\log(a^{-n}) = -n\log(a)$.

If $r=\frac{p}{q}$ is a rational, with $q\gt 0$, then we have that $$q\log(a^{p/q}) = \log(a^{p}) = p\log(a),$$ hence $\log(a^{p/q}) = \frac{p}{q}\log(a)$.

Now, since $\log(x)$ is strictly increasing, it has an inverse. Let us denote the inverse by $E$. So $E(x)=y$ if and only if $\log(y)=x$. From the properties of $\log(x)$, we get:

1. $E(x)$ is strictly increasing and differentiable.
2. Since $x=\log(E(x))$, differentiating both sides and using the Chain Rule we have $$\begin{align*} 1 &= \frac{1}{E(x)}\left(\frac{d}{dx}E(x)\right)\\ E(x) &= \frac{d}{dx}E(x). \end{align*}$$
3. $\lim_{x\to-\infty}E(x) = 0^+$ (that is, $E(x)$ is positive for all $x$, and approaches $0$ as $x\to-\infty$), and $\lim_{x\to\infty}E(x)=\infty$.
4. $E(0)=1$.

Now, say $E(a)=r$ and $E(b)=s$. Then $\log(r)=a$ and $\log(s)=b$. Therefore,$a+b=\log(r)+\log(s) = \log(rs)$ and $a-b=\log(r)-\log(s)=\log(r/s)$, hence $$E(a+b) = rs = E(a)E(b)\quad\text{and}\quad E(a-b) = \frac{E(a)}{E(b)}.$$ giving 5. $E(a)E(b) = E(a+b)$. 6. $\frac{E(a)}{E(b)} = E(a-b)$.

If $E(a)=r$ and $n\gt 0$ is an integer, then by induction we have that $E(na) = E(a)^n = r^n = E(na)$. This immediately extends to negative integers, and finally to rational exponents, so that $E(a)^{p/q} = E(ap/q)$ for all integers $p,q$, with $q\gt 0$.

Definition. If $a\gt 0$ and $b$ is any real number we define $\exp(a,b) = E(b\log(a))$.

Proposition. If $b$ is a rational number, then $\exp(a,b)=a^b$.

Proof. First, assume $b$ is a positive integer. By induction: $\exp(a,1) = E(1\log(a)) = E(\log(a)) = a = a^1$. If $\exp(a,k) = a^k$, then $$\exp(a,k+1) = E((k+1)\log(a)) = E(k\log(a)+\log(a)) = E(k\log(a))E(\log(a)) = a^ka = a^{k+1}.$$ Thus, the proposition holds for $b$ a positive integer.

If $b$ is a negative integer, then $$\exp(a,b) = E(b\log(a)) = E(-(-b)\log(a)) = \frac{1}{E((-b)\log(a))} = \frac{1}{a^{-b}} = a^b.$$

Finally, if $b=\frac{p}{q}$ with $p,q$ integers and $q\gt 0$, then $$(\exp(a,b))^q = (E(b\log(a)))^q = E(qb\log(a)) = E(p\log(a)) = a^p,$$ so taking $q$th roots on both sides we get $\exp(a,b) = a^{p/q}=a^b$. $\Box$

In light of the proposition, we simply write $a^b$ instead of $\exp(a,b)$; we know it agrees with the "old" definition when $b$ is a rational number.

Thus, if $\{q_n\}_{n\in\mathbb{N}}$ is a sequence of rational numbers and $q_n\to r$, then $$\lim_{n\to\infty} a^{q_n} = \lim_{n\to\infty}E^{q_n\log(a)} = E\left(\lim_{n\to\infty}q_n\log(a)\right) = E(r\log(a)) = a^r,$$ since $E$ is continuous. Thus, this definition agrees with the definition you propose for $a^r$.

So, finally, using the Chain Rule, we get $$\begin{align*} \frac{d}{dx}x^r &= \frac{d}{dx} E(r\log(x)) \\ &= E(r\log(x))\left(\frac{d}{dx}r\log(x)\right) \\ &= \frac{r}{x}E^(r\log(x))\\ &= rx^{-1}E(r\log(x))\\ &= rE(-\log(x))E(r\log(x))\\ &= rE(-\log(x)+r\log(x)) = rE((r-1)\log(x))\\ &= rx^{r-1}, \end{align*}$$ giving the desired differentiation formula.

Of course, we let $e$ be the solution to $\log(x)=1$, and then we have that $$E(x) = E(x\log(e)) = e^x,$$ which gives us the "usual" notation. Then Taylor's Theorem will give the equivalence of this definition with the power series definition.