Why is $\log_e(x)'=\ln(x)'= 1/x$? how to prove the differentiation rule. ofc that means also $\int(1/x)dx=\ln(x) + c$
I think that the derivative of the $\ln$ of a variable with respect to the variable being 1/x is an axiom to prove that the limit of
$(1 + 1/(ax))^{bx}$ as $x \to \infty$ is $e^{b/a}$
How to prove one without using the other and any theorem following from them ?
Or at least can we prove one of them without the other and the other should follow?
It certainly is not an axiom.
If $\ln$ is defined as the inverse of the exponential function (which is its own derivative): $$y=\ln x\quad(x>0)\overset{\text{def}}{\iff} x=\mathrm e^y,$$ then $$(\ln)'(x)=\frac 1{(\mathrm e^y)'}=\frac 1{\mathrm e^y}=\frac 1x.$$
doing math, namely quotients and exponents?
– Bernard
Oct 07 '17 at 23:32
$ ... $ for inline maths, between $$ ... $$ for displayed formulae, or one of the amsmath environments for multiline displayed environments. For instance $a:b$ as a true fraction is coded \frac{a}{b} and so on. I think there are tutorials on this site.
– Bernard
Oct 07 '17 at 23:58
That $(\ln x)'=\frac1x$ follows from the definition of $\ln x$ and the fundamental theorem of calculus. .. $\ln x=\int_1^x \frac 1t dt$ by definition. ..
It's possible to define $\ln(x)$ or $e$ in numerous different ways, such that they don't have to depend on each other. So too can you prove that $\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\frac{1}{x}$ or $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$, though the method may depend on how you've defined each construct.
@Bernard's answer covers the case that $\ln(x):=y$ such that $x=e^y$. Likewise, @Chris Custer's answer covers the case that $\ln(x):=\int_1^x\frac{1}t\,\mathrm{d}t$. For completeness, there's a few other ways you can define $\ln(x)$.
Limit Definition
If $\ln(x):=\lim_{h\to0}\frac{x^h-1}{h}$, then:
$$\begin{aligned}\frac{\mathrm{d}}{\mathrm{d}x}\ln(x) &=\lim_{g\to0}\frac{\lim_{h\to0}\frac{(x+g)^h-1}{h}-\lim_{h\to0}\frac{x^h-1}{h}}{g} \\\text{Definition of derivative.}& \\&=\lim_{g\to0}\lim_{h\to0}\frac{(x+g)^h-1-x^h+1}{hg} \\\text{Multiply through by $h$.}& \\&=\lim_{g\to0}\lim_{h\to0}\frac{x^h(\frac{g}{x}+1)^h-x^h}{hg} \\\text{Put in form $(z+1)^\alpha$.}& \\&=\lim_{g\to0}\lim_{h\to0}\frac{x^h\left[1+h\frac{g}{x}+\binom{h}{2}\frac{g^2}{x^2}+\ldots\right]-x^h}{hg} \\\text{Binomial Series.}& \\&=\lim_{g\to0}\lim_{h\to0}\frac{\color{red}{\not}{x^h}+hgx^{h-1}+\binom{h}{2}g^2x^{h-2}+\ldots-\color{red}{\not}{x^h}}{hg} \\\text{Cancel terms.}& \\&=\lim_{g\to0}\lim_{h\to0}\left[x^{h-1}+hg(\ldots)\right] \\\text{Factor $h$ and $g$.}& \\&=x^{-1} \end{aligned}$$
You may ask how we'd calculate $y=x^{0.0\ldots01}$ without logarithms. We can do it with the Newton-Raphson method, finding the roots of $y^{100\ldots0}-x=0$. We can use similar methods to compute $x^r$ for rational $r$.
Series Definition
We can also define $\ln(x)$ by its Taylor series. In this case, its derivative is, for $0<x<2$:
$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\ln(x) &=\frac{\mathrm{d}}{\mathrm{d}x}\left(-\sum_{n\geq1}\frac{(1-x)^n}{n}\right) \\&=-\left(\sum_{n\geq1}\frac{n(1-x)^{n-1}\cdot(-1)}{n}\right) \\\text{Differentiate each term.}& \\&=\sum_{n\geq1}(1-x)^{n-1} \\\text{Cancelled $n$ and $-1$.}& \\&=\frac{1}{1-(1-x)} \\\text{Limit of geometric series.}& \\&=\frac{1}{x} \end{aligned} $$
Whether it's really justified to differentiate an infinite series is a bit hand-wavey, I'll admit but I think this answer gives a fairly good feel for how you can show that $\frac{\mathrm{d}}{\mathrm{d}x}\ln(x)=\frac{1}{x}$ with alternative definitions. In fact we haven't used $e$ once. We can probaly be able to prove the derivative with other means too. This Question gives some ways of proving it.
