-3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Do you have access to the theorem which tells you something about the derivative of the inverse function? I.e. using that $\ln(x)$ is the inverse of $\exp(x)$ (and that $\exp'(x)=\exp(x)$). $\endgroup$ – Roland Oct 7 '17 at 21:37
  • 1
    $\begingroup$ Euler's constant $\gamma \approx 0.577...$ is something different from $e \approx 2.718...$. $\endgroup$ – Jean Marie Oct 7 '17 at 21:42
  • $\begingroup$ @JeanMarie People often to refer to $e$ as Euler's constant too. Besides, the context makes it clear which is being referred to. $\endgroup$ – Simply Beautiful Art Oct 7 '17 at 21:42
  • 1
    $\begingroup$ "I think... is an axiom". Actually, all of these and many others are actually equivalent definitions (not axioms) since given any one of them, you can prove the other. However, you'll need to tell us what your definitions are. How do you define $\ln(x)$? How do you define $e^x$? $\endgroup$ – Simply Beautiful Art Oct 7 '17 at 21:57
  • 1
    $\begingroup$ @SimplyBeautifulArt: I think $\mathrm e$ is referred to as Euler's number, not ‘Euler's constant’. $\endgroup$ – Bernard Oct 7 '17 at 22:08
2
$\begingroup$

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 $$(\ln)'(x)=\frac 1{(\mathrm e^y)'}=\frac 1{\mathrm e^y}=\frac 1x.$$

$\endgroup$
  • $\begingroup$ You used what you want to prove. $\endgroup$ – George Ntoulos Oct 7 '17 at 22:28
  • $\begingroup$ @GeorgeNtoulos: How that? $\endgroup$ – Bernard Oct 7 '17 at 22:37
  • $\begingroup$ You are using e. Isn't what you said implicit I said to prove one without using the other and any theorem following from them ! $\endgroup$ – George Ntoulos Oct 7 '17 at 22:48
  • $\begingroup$ This depends on the definition you have for the exponential. The standard definition is it's the sum of the power series $\;1+x+\dfrac{x^2}2+\dots+\dfrac{x^n}{n!}+\dotsm$. Almost equivalent: you can say it is the solution of the differential equation $y'=y$ with initial condition $y(0)=1$. $\endgroup$ – Bernard Oct 7 '17 at 23:05
  • $\begingroup$ 1st question how do you do math quotients and exponents. $\endgroup$ – George Ntoulos Oct 7 '17 at 23:14
0
$\begingroup$

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. ..

$\endgroup$
  • $\begingroup$ From the way the OP has posed the question, it looks like they aren't taking $\ln(x)=\int_1^x\frac1t~\mathrm dt$ by definition. $\endgroup$ – Simply Beautiful Art Oct 7 '17 at 22:14
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The problem I have is that we use lns property of ln(x)'=1/x to calculate (1 + 1/x)^x as x aproaches inf. And we use e=that limit to prove that ln(x)'=1/x integrals inverse functions et al are implicitly using the above when e^x'=e'x then ln(x)'=1/x $\endgroup$ – George Ntoulos Oct 8 '17 at 0:09
  • $\begingroup$ @GeorgeNtoulos If you mean to say that the definitions are circular, then that's what I've been trying to show isn't necessarily true. We can chose numerous different ways of defining $e^x$ and $\ln(x)$ such that we are doing logically sound steps. Case in point, we could define $e=\sum_{n\geq0}\frac{1}{n!}$. $\endgroup$ – Jam Oct 8 '17 at 0:13
  • $\begingroup$ We also don't have to use $\ln$ at all in calculating $\lim_{x\to+\infty}(1+1/x)^x$. We can show that it converges to a finite value and define that value as $e$. $\endgroup$ – Jam Oct 8 '17 at 0:15
  • $\begingroup$ The Mathematics beats me $\endgroup$ – George Ntoulos Oct 8 '17 at 0:21
  • $\begingroup$ @GeorgeNtoulos You'll get there, if you stick at it. You're clearly asking the right questions. It's a good thing to question the definitions of particular mathematical constructs, but in this case, $\ln$ and $e$ are very well established so they've got quite a solid framework :) $\endgroup$ – Jam Oct 8 '17 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.