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

Question

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:

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

But I didn't find a way to get something useful out of this. Please help me.

Answer 1

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...

Answer 2

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 $\log'{x}$.

Answer 3

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

Answer 4

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 $\log'(x)=1/x$.