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Let the function $f$ be defined as: $f(x)=x^r, r\in \mathbb R$.

We have all heard that $f'(x)=r x^{r-1}$. And apparently, there exists a proof for that. But in that proof, there is something that doesn't look convincing to me. As you see, we first take this into account: $$f(x)=x^r=e^{r\ln{x}}$$ and then, use the existing rules for the derivative of exponential functions.

But let's take a little step back. As far as I know, the exponential function is defined as the inverse of the logarithm function. And the logarithm is defined as: $$\ln(x)=\int_1^x \frac{1}{t} dt,\quad x>0$$

At the first steps, it is proved that this function is well-defined, continuous and anywhere differentiable in its domain. Then we prove that it is injective, and denote its inverse function by $\exp(x)$ and we don't mention anything about the term exponential. Then it is shown that the derivative of this particular function is equal to itself, by using those abstract definitions and some smart-looking workarounds.

But something caught my attention that, in the process of those proofs, it is taken as an assumption that the derivative of the power function is equal to $rx^{r-1}$, and this seems like a loop to me. Also if the other definition of the $\exp$ is used, which is based on the Taylor series, we would still be stuck to the derivative of power function.

So, is there a way to prove the derivative of the power function other than using the exponential function? Or in other words, is there any way to differentiate the so-called $\exp$ without using the power function?

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    $\begingroup$ You say that "in the process of those proofs" the assumption that the derivative of $x^r $ is $rx^{r-1} $. I don't think it is, so it would be nice if you quote the proofs you are talking about. $\endgroup$ – Martin Argerami Aug 30 '16 at 6:11
  • $\begingroup$ @MartinArgerami I read it a while ago, and I am searching now $\endgroup$ – polfosol Aug 30 '16 at 6:12
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    $\begingroup$ It may be noted that there is an elementary proof (using induction) that for every positive integer $n$, the derivative of $x\mapsto x^n$ is $x\mapsto nx^{n-1}$; this is independent of anything about exponential functions. It is not quite clear to me where you think to see a circularity in the proofs, but it might be that what is used there is only this integral power case, in which case the circularity vanishes. $\endgroup$ – Marc van Leeuwen Aug 30 '16 at 6:17
  • $\begingroup$ I came across this, theorems after #22 are discussing the above-mentioned procedure. Although all it talks about is rational numbers and there is no mention of real numbers and thus no circularity, but I am somehow sure that I have seen it somewhere $\endgroup$ – polfosol Aug 30 '16 at 6:43
  • $\begingroup$ re "is there a way to prove the derivative of the power function other than using the exponential function?" You defined the power function via the exponential function. $\endgroup$ – quid Aug 30 '16 at 7:55
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The linked pdf (see comments to question by OP) gives a theory of exponential, logarithmic and general power functions of a real variable using the simplest possible approach (here simple means short and concise proofs and we assume that the theory of Riemann integrals is already developed and available for use) and this approach of defining logarithm via integral is very common in many textbooks of calculus/analysis.

There is no circularity involved in the proof of the derivative formula $$\frac{d}{dx}(x^{r}) = rx^{r - 1}\tag{1}$$ for $x > 0$ and $r \in \mathbb{R}$. It should however be noted that when $r$ is rational then the function $f(x) = x^{r}$ can be defined without any reference to $\log$ or $\exp$ and a proof of derivative formula then follows from algebraic inequalities. Moreover it is possible to extend this definition to define $x^{r}$ for any irrational $r$ and then the derivative formula $(1)$ also holds according to this extended definition. See this answer for more details on this approach which avoids logarithmic/exponential functions as far as the definition of $x^{r}$ and calculation of its derivative are concerned.

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  • $\begingroup$ The linked answer was very helpful. Thank you $\endgroup$ – polfosol Aug 30 '16 at 10:11
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    $\begingroup$ I don't agree using integrals is simplest possible (a lot needs to be set up before one can use integrals), but your linked answer is excellent and is a great example of just how powerful the squeeze theorem can be; a lot of such theorems can be proven in this way, essentially by delaying the limit computation until the end we can manipulate bounds with no issues with swapping limits. $\endgroup$ – user21820 Aug 30 '16 at 12:18
  • $\begingroup$ @user21820: I agree that we need to develop the theory of Riemann integrals and this itself is heavy machinery. But if we assume this machinery then the definition of logarithm as integral gives the properties of $\log,\exp$ easily. I wanted to highlight this fact. Perhaps you prefer the infinite series for $\exp$. $\endgroup$ – Paramanand Singh Aug 30 '16 at 15:46
  • $\begingroup$ @ParamanandSingh: I'm slightly in favour of the Taylor series for the complex-valued $\exp$, because it is easy to prove in one go all the properties, which then immediately yield the important properties of complex-valued trigonometric functions too. Otherwise it can be a bit painful to set up; one of my teachers used the real-valued $\exp$ to define $\exp(x+yi) = \exp(x) ( \cos(y) + i \sin(y) )$ for real $x,y$, and then proved that $\exp$ is complex-differentiable! No motivation! Of course it is clear if we went the other way, but that suggests this way isn't the 'right' way. =) $\endgroup$ – user21820 Aug 30 '16 at 16:04
  • $\begingroup$ @user21820: All I can say is that the basic theory of infinite series (as much as needed for series for $\exp(z)$) is easier to develop compared to a proper theory of Riemann integrals. And therefore the approach of Taylor's series is even easier and general enough to deal with complex numbers. But I have found the definition $\lim_{n \to \infty}(1 + (z/n))^{n}$ for complex $z$ also useful. You may have a look at math.stackexchange.com/a/1668179/72031 $\endgroup$ – Paramanand Singh Aug 30 '16 at 16:10
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$$\newcommand{\dd}{\frac{\mathrm{d}}{\mathrm{d}x}}$$

I haven't looked at your link, but it's important to note that the rule for taylor series and the rule you want to prove are not the same.

The first one says that if $r\in\mathbb R$, then $\dd x^r = rx^{r-1}$. For the taylor series rule, we only need that $r\in\mathbb N$, then $\dd x^r = rx^{r-1}$. This is much easier to prove (namely, we can use the difference quotient). Calculations with the difference quotient show that: \begin{align} \dd x^r & = \lim_{h\to 0}\frac{(x+h)^r-x^r}{h} \end{align} We have the binomial formula that $$(a+b)^n = \sum_{i = 0}^n \binom{n}{i}a^ib^{n-i}$$ We can use this to find that: \begin{align} \dd x^r & = \lim_{h\to 0}\frac{\sum_{ i = 0}^r\binom{r}{i}x^ih^{r-i}-x^r}{h} \\ &=\lim_{h\to 0}\frac{h\binom{r}{r-1}x^{r-1}+h^2\sum_{i = 0}^{r-2}\binom{r}{i}x^{i}h^{r-2-i}}{h} \\ &=\binom{r}{r-1}x^{r-1}=\frac{r!}{(r-1)!}x^{r-1} = rx^{r-1} \end{align}

As we used the binomial formula, we assume that $r\in\mathbb N$. In this way, it wasn't a strong enough proof to establish the result you want. It WAS strong enough to establish the power rule for differentiation taylor series. In this way, the proof isn't circular.

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  • $\begingroup$ This was exactly my first thought. $\endgroup$ – Andres Mejia Aug 30 '16 at 6:04
  • $\begingroup$ This seems legit. But what about the other definition of $\exp$? $\endgroup$ – polfosol Aug 30 '16 at 6:06
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    $\begingroup$ We can use the inverse function theorem to establish the derivative of the inverse of a monatonic function (which $\ln x$ is). See page 299 of this for a proof. The proof doesn't rely on any prior rules of differentiation, just the difference quotient. This book also defines$\exp$ as the inverse of $\ln$, which you may be interested in. $\endgroup$ – Mark Aug 30 '16 at 6:11
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I see no such circular logic in the link you provided. The proof there only uses the Chain Rule and the derivative of $\ln(x)$ as established facts outside the scope of the proof. It reaches a point where it asks the reader to compute $$\frac{d}{dx}e^{r\ln(x)}$$ and then uses the Chain Rule, the derivative of $\ln(x)$, and the derivative of $e^x$ (established earlier in the proof) to calculate this derivative is $$r x^{r-1}\text{.}$$

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Here's a way to prove it at least for rational exponents without resorting to the exponential function or logarithms. Extending to real coefficients probably can be done by using a sequence of rational numbers $q_n$ converging to the $r$, but then one has to show that the limits for $n\to\infty$ and $h\to 0$ can be exchanged, which I don't know how to do.

In the following, for simplicity I assume that $x>0$; the case $x\le 0$ needs additional considerations of whether the functions are defined, and when the power laws can be used.

First, use induction and the product rule (which I assume has already been proven before) to prove it for $n\in\mathbb N$.

Obviously, it's true for $n=0$: $(x^0)' = 1' = 0 = 0x^{-1}$

Assume it is true for $x^n$ Then we find: $$(x^{n+1})' = (x\cdot x^n)' = x'\cdot x^n + x\cdot (x^n)' = x^n + x\cdot (nx^{n-1}) = (n+1)x^n$$

So it's true at least for $x\in\mathbb N$.

We can also determine that it is true for $x^{-1}$ by explicitly calculating the limit: $$\lim_{h\to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h} = \lim_{h\to 0}\frac{-1}{x(x+h)}= -\frac{1}{x^2}$$

Now we can use the chain rule (which I also assume to have been proven) to extend the rule to negative integers, and thus to all of $\mathbb Z$: $$(x^{-n})' = ((x^n)^{-1})' = -\frac{1}{(x^n)^2}\cdot (x^n)' = -n\frac{x^{n-1}}{x^{2n}} = -nx^{-n-1}$$

Next, we can use the fact that $x^{1/n}$ is the inverse function of $x^n$ and use the rule of the derivative of inverse functions (also assumed to already have been proved) to get $$(x^{\frac{1}{n}})' = \frac{1}{n(x^{1/n})^{n-1}} = \frac{1}{n}x^{\frac{1}{n}-1}$$

Now we can use the chain rule again for $x^{m/n} = (x^m)^{1/n}$ to prove the rule for arbitrary rational exponents.

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  • $\begingroup$ This looks interesting $\endgroup$ – polfosol Aug 30 '16 at 9:06
  • $\begingroup$ But for applying the chain rule, isn't it necessary for a function to have inverse and hence be injective? $x^2$ doesn't look like injective to me. $\endgroup$ – polfosol Aug 30 '16 at 9:09
  • $\begingroup$ @polfosol: Note that at the very beginning, I restricted to $x>0$. For $x>0$ the power function is injective. Also, the common definition of $\sqrt{x}=x^{1/2}$ takes exactly this branch. $\endgroup$ – celtschk Aug 30 '16 at 9:13
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Not the proper 'expansion' proof but a short and also sweet one.

Let $$y=x^n, $$

For a small change $dx$ in $x$ let there be a change $dy$ in value of $f(x)$ (which is equal to $y$, in our case).

$$y+dy=(x+dx)^n$$ $$=x^n(1+\frac{dx}{x})^n$$

As $\frac{dx}{x}$ is much smaller than $ 1$,

$$=x^n(1+n\frac{dx}{x})$$ $$y+dy=x^n+nx^{n-1}dx$$

Now $y=x^n$, $$dy=nx^{n-1}dx$$ Which gives $$\frac{dy}{dx}=f'(x)=nx^{n-1}.$$

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    $\begingroup$ I think the part $(1+\frac{dx}{x})^n \approx (1+n \frac{dx}{x})$ needs a little more work $\endgroup$ – polfosol Aug 30 '16 at 7:39
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    $\begingroup$ @polfosol plz explain why fr i encounter the same problem evry time. $\endgroup$ – Display name Aug 30 '16 at 8:09

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