# Please help me to show, that $(\ln x)'=\frac1 x$

In school, we recently started with derivations. I looked into a list of simple derivations and tried to prove them, in order to practice. Now, I tried to find the derivative of $\ln x$, but I got stuck. Some web pages suggest to use the identity $e = \lim_{h\to\infty}\left(1+h^{-1}\right)^h$, but I still don't get a solution. I started by the basic approach:

$$(\ln x)'=\lim_{\Delta\to0}\frac{\ln(x+\Delta)-\ln x}\Delta$$

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Is your last displayed equation supposed to have $(\ln x)'$ on the left hand side? Otherwise, it's incorrect. – Arturo Magidin Apr 6 '11 at 19:21
Use de l'Hôpital ;-) (sorry bad joke) – Fabian Apr 6 '11 at 19:21
@Arturo Magidin: Yes. Sorry. Got confused by myself. – FUZxxl Apr 6 '11 at 19:22
@Fabian: In our school, we are doing l'Hôpital in grade 11, I don't want to wait that long ;) – FUZxxl Apr 6 '11 at 19:23
Instead of taking the derivative of of $y = \ln x$, might you take the derivative of $x = \exp y$? – Henry Apr 6 '11 at 19:24

The simplest way to find the derivative of the natural logarithm is to use the Inverse Function Theorem (or the Chain Rule), but since you say you only recently started, you may not know it yet.

So instead, we begin with two ingredients. One is that $\ln(u)$ is continuous. That means that if $\lim\limits_{x\to a}f(x)$ exists, then $$\lim_{x\to a}\ln(f(x)) = \ln\left(\lim_{x\to a}f(x)\right).$$

The second ingredient (which you may or may not know yet) is that $$\lim_{h\to\infty}\left(1 + \frac{a}{h}\right)^h = e^a.$$ To see this, note that this is immediate if $a=0$; if $a\gt 0$, then just do a quick rewrite: \begin{align*} \lim_{h\to\infty}\left(1 + \frac{a}{h}\right)^h &= \lim_{h\to\infty}\left( 1 + \frac{1}{(h/a)}\right)^h\\ &=\lim_{h\to\infty}\left(\left(1 + \frac{1}{(h/a)}\right)^{h/a}\right)^a\\ &= \left(\lim_{h\to\infty}\left(1 + \frac{1}{(h/a)}\right)^{h/a}\right)^a. \end{align*} If $a\gt 0$, then $h/a\to\infty$ as $h\to\infty$, so by the definition of $e$ you get that $$\lim_{h\to\infty}\left(1+\frac{a}{h}\right)^h = \left(\lim_{(h/a)\to\infty}\left(1 + \frac{1}{(h/a)}\right)^{h/a}\right)^a = (e)^a = e^a.$$ If $a\lt 0$, then replacing $a$ with $-a$ we can do the same trick as above after proving that $$\lim_{h\to\infty}\left(1 - \frac{1}{h}\right)^h = e^{-1}.$$ Indeed, though it takes a bit more algebraic trickery: \begin{align*} \lim_{h\to\infty}\left(1 - \frac{1}{h}\right)^h &= \lim_{h\to\infty}\left(\frac{h-1}{h}\right)^h = \lim_{h\to\infty}\left(\frac{h}{h-1}\right)^{-h}\\ &= \left(\lim_{h\to\infty}\left(\frac{(h-1)+1}{h-1}\right)^h\right)^{-1}\\ &= \left(\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)^h\right)^{-1}\\ &=\left(\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)^{h-1}\left(1 + \frac{1}{h-1}\right)^1\right)^{-1}\\ &= \left(\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)^{h-1}\lim_{h\to\infty}\left(1 + \frac{1}{h-1}\right)\right)^{-1}\\ &= \Bigl((e)(1)\Bigr)^{-1} = e^{-1}.\end{align*}

Then, in the previous limit, if $a\lt 0$ then replace it with $-a$ and change the $+$ to a $-$, to get that the limit equals $(e^{-a})^{-1} = e^a$ as well.

And finally, with these ingredients in hand, we are ready. We have: \begin{align*} \frac{d}{dx}\ln x &= \lim_{\Delta\to 0}\frac{\ln(x+\Delta)-\ln(x)}{\Delta}\\ &= \lim_{\Delta\to 0}\frac{1}{\Delta}\left(\ln(x+\Delta)-\ln(x)\right)\\ &=\lim_{\Delta\to 0}\frac{1}{\Delta}\ln\left(\frac{x+\Delta}{x}\right)\\ &=\lim_{\Delta\to 0}\frac{1}{\Delta}\ln\left(1 +\frac{\Delta}{x}\right)\\ &=\lim_{\Delta\to 0}\ln\left(\left(1 + \frac{\Delta}{x}\right)^{1/\Delta}\right)\\ &= \lim_{\Delta\to 0}\ln\left(\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta}\right). \end{align*} If $\Delta\to 0^+$, then $\frac{1}{\Delta}\to\infty$, so letting $h=\frac{1}{\Delta}$ we have: $$\lim_{\Delta\to 0^+}\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta} = \lim_{h\to\infty}\left( 1 + \frac{1/x}{h}\right)^h = e^{1/x}.$$ If $\Delta\to 0^-$, then $\frac{1}{\Delta}\to-\infty$, so letting $h=-\frac{1}{\Delta}$, we have: \begin{align*} \lim_{\Delta\to 0^-}\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta} &= \lim_{h\to\infty}\left(1 - \frac{1/x}{h}\right)^{-h}\\ &= \lim_{h\to\infty}\left(\left(1 - \frac{1/x}{h}\right)^{h}\right)^{-1}\\ &= \left(e^{-1/x}\right)^{-1} = e^{1/x}. \end{align*} Therefore, we have: \begin{align*} (\ln x)' &= \lim_{\Delta\to 0}\frac{\ln(x+\Delta)-\ln(x)}{\Delta}\\ &= \ln\left(\lim_{\Delta\to 0}\left(1 + \frac{1/x}{1/\Delta}\right)^{1/\Delta}\right)\\ &= \ln\left(e^{1/x}\right) = \frac{1}{x}. \end{align*}

And this is why the Chain Rule or the Inverse Function Theorem are such a better way of proving this...

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Thank you. Great answer. – FUZxxl Apr 6 '11 at 20:11

Another definition is that $\log$ is the inverse function of $\exp$. With this definition, you have $\log(\exp(t))=t$. By the chain rule, $\log'(\exp(t))\exp'(t)=1$ and so $\log'(\exp(t))=1/\exp(t)$. Write $x=\exp(t)$ and get $\log'(x)=1/x$.

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It's not difficult $$\lim\limits_{\delta\to 0}\frac{\log(x+\delta) - \log{x}}{\delta} = \lim\log\left(1 - \frac{\delta}{x}\right)^{1/\delta}$$

and then just put $h = \frac{1}{\delta}$. Btw, you have a typo $(e^x)'$ - you better write $\log'{x}$.

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Ah... Okay. And how to show, that $\lim_{\delta\to\infty}\ln\left(1-\frac1{x\delta}\right)^\delta = x^{-1}?$ – FUZxxl Apr 6 '11 at 19:27
@FUZxxl: First, you only get $\frac{1}{\delta}\to\infty$ if $\delta\to 0^+$, so you'll need to do the two one-sided limits separately. Second: show that $\lim_{h\to\infty}(1 + \frac{k}{h})^{h} = e^k$. Third: Remember that the logarithm is continuous, so $\lim\log(g) = \log(\lim(g))$. – Arturo Magidin Apr 6 '11 at 19:32
@FUZxxl: maybe this link math.stackexchange.com/questions/12307/… helps? – Fabian Apr 6 '11 at 19:33
@Arturo Magidin: Thank you very much. – FUZxxl Apr 6 '11 at 19:41

The definition is $$\log(x) = \int_1^x \frac{1}{t} \mathrm{d}t$$ and inverting it by the fundamental theorem of calculus gives $$\frac{\mathrm{d}}{\mathrm{d}x} \log(x) = \frac{1}{x}.$$

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That's a definition of the natural log; and since the OP says he only recently started with derivations, how high do you estimate the likelihood the OP is familiar with the FTC? – Arturo Magidin Apr 6 '11 at 19:25
I always thought the definition is that $\log$ is the inverse function of $\exp$. – Fabian Apr 6 '11 at 19:26
If he doesn't know it maybe he will find my post interesting and learn it. – quanta Apr 6 '11 at 19:26
I think that although this may not be the most useful response for the OP, it is one of the most direct derivations. Any derivation of $log'$ requires a definition of either $log$ or $exp$, and this is a pretty good one. IIRC, this is how it's done in Spivak's Calculus, and then $exp'$ is derived by brute force. – Kopper Apr 6 '11 at 20:16
"Is it a definition?" - a bit like asking which of the chicken or the egg came first. One can start the theory with either of $\exp$ or $\ln$ and derive everything else from there... – J. M. Apr 7 '11 at 6:36