39
$\begingroup$

I'm trying to prove that $\frac{\mathrm{d} }{\mathrm{d} x}\ln x = \frac{1}{x}$.

Here's what I've got so far: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\ &= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\ &= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\ \end{align} $$ To simplify the logarithm: $$ \lim_{h\to0}\left (1 + \frac{h}{x}\right )^{\frac{1}{h}} = e^{\frac{1}{x}} $$ This is the line I have trouble with. I can see that it is true by putting numbers in, but I can't prove it. I know that $e^{\frac{1}{x}} = \lim_{h\to0}\left (1 + h \right )^{\frac{h}{x}}$, but I can't work out how to get from the above line to that. $$ \lim_{h\to0}\left ( \left (1 + \frac{h}{x}\right )^{\frac{1}{h}}\right )^{h} = e^{\frac{h}{x}} $$ Going back to the derivative: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(e^{\frac{h}{x}})}{h} \\ &= \lim_{h\to0} \frac{\frac{h}{x}\ln(e)}{h} \\ &= \lim_{h\to0} \frac{h}{x} \div h\\ &= \frac{1}{x} \\ \end{align} $$

This proof seems fine, apart from the middle step to get $e^{\frac{1}{x}}$. How could I prove that part?

$\endgroup$
6
  • 16
    $\begingroup$ Which definition of the natural logarithm do you have? $\endgroup$
    – Bernard
    Jun 28, 2015 at 10:29
  • 1
    $\begingroup$ $\ln$ is continuous, so you can say $\lim_n \ln x_n = \ln \lim_n x_n$, and the other thing is just one definition of $e^x$. What is your problem exactly, can you elaborate a bit? $\endgroup$
    – krvolok
    Jun 28, 2015 at 10:30
  • $\begingroup$ @Bernard I have the natural logarithm defined as the inverse of $e^{x}$. $\endgroup$
    – rlms
    Jun 28, 2015 at 10:35
  • $\begingroup$ @krvolok My problem is the line shown - I can't algebraically work out how to prove it. I'll edit the question to include more details. $\endgroup$
    – rlms
    Jun 28, 2015 at 10:37
  • $\begingroup$ See also: proofwiki.org/wiki/Derivative_of_Natural_Logarithm_Function $\endgroup$
    – user845875
    Sep 28, 2021 at 12:33

5 Answers 5

53
$\begingroup$

If you can use the chain rule and the fact that the derivative of $e^x$ is $e^x$ and the fact that $\ln(x)$ is differentiable, then we have:

$$\frac{\mathrm{d} }{\mathrm{d} x} x = 1$$

$$\frac{\mathrm{d} }{\mathrm{d} x} e^{\ln(x)} = e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$

$$e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$

$$x \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$

$$\frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = \frac{1}{x}$$

$\endgroup$
4
  • 15
    $\begingroup$ Note, however, that this assumes that $\ln x$ is differentiable. (That is required if you want to use the chain rule) So unless you have proved that $\ln x$ is differentiable, this proof cannot work. As far as I can see, there is no better way to prove that $\ln x$ is differentiable that to calculate the derivative explicitly. $\endgroup$ Jun 28, 2015 at 12:07
  • 1
    $\begingroup$ @cmtappu96 Indeed, good point. $\endgroup$
    – wythagoras
    Jun 28, 2015 at 12:13
  • 32
    $\begingroup$ The inverse function theorem guarantees that $\ln x$ is differentiable. $\endgroup$
    – mweiss
    Jun 28, 2015 at 16:33
  • 1
    $\begingroup$ The antiderivative of $\frac 1 x$ is $\ln |x|$, however. $\endgroup$ Aug 2, 2021 at 16:32
30
$\begingroup$

The simplest way is to use the inverse function theorem for derivatives:

If $f$ is a bijection from an interval $I$ onto an interval $J=f(I)$, which has a derivative at $x\in I$, and if $f'(x)\neq 0$, then $f^{-1}\colon J\to I$ has a derivative at $y=f(x)$, and $$\bigl(f^{-1}\bigr)'(y)=\frac1{f'(x)}=\frac1{f'\bigl(f^{-1}(y)\bigr)}.$$

As $(\mathrm e^x)'=\mathrm e^x\neq 0\,$ for all $x$, we know that $\,\ln\,$ has a derivative at each point of its domain, and $$(\ln)'(y)=\frac1{\mathrm e^{\,\ln y}}=\frac1y.$$

$\endgroup$
17
$\begingroup$

Define $$e=\lim_{h\to 0} \left(1+h\right)^{1/h}.$$ Then change variables $h\mapsto h/x$ giving $$e=\lim_{h/x\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}}=\lim_{h\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}},$$ where the limit in the second equality follows since $h$ approaches $0$ as $h/x$ does. Since $x$ is constant w.r.t. $h$, we can simplify by raising both sides to the power $1/x$, giving you the desired identity.

$\endgroup$
7
$\begingroup$

Just throwing it out there for you to see, I also like this proof:

$$y=\ln x$$ $$e^y=x$$

after differentiating,

$$e^y \frac{dy}{dx}=1$$

$$ \begin{align} \frac{dy}{dx}&=\frac{1}{e^y}\\ &= \frac{1}{e^{\ln x}}\\ &= \frac{1}{x} \end{align} $$

of course, that assumes you already know the derivative of $e^x$ and the chain rule

$\endgroup$
2
  • 3
    $\begingroup$ For this method, you also need to know that $\: \ln \:$ is differentiable. $\;\;\;\;$ $\endgroup$
    – user57159
    Jun 28, 2015 at 23:10
  • $\begingroup$ If $y=e^x$ is differentiable and its derivative is not $0$ where you are looking, then $x=\ln(y)$ is differentiable. Inverse Function Theorem $\endgroup$
    – robjohn
    Oct 13, 2021 at 22:14
2
$\begingroup$

If you can use the definition of $e$ as: $$e:=\lim_{n\rightarrow∞}\left(1+\frac{1}{n}\right)^n$$

and the slightly modified form: $\displaystyle e^x=\lim_{n\rightarrow∞}\left(1+\frac{x}n\right)^n$

then, by setting $h=\frac1{x}$ you can calculate the desired limit.

$\endgroup$
3
  • 1
    $\begingroup$ How do you get from the definition to the slightly modified form? $\endgroup$
    – rlms
    Jun 28, 2015 at 11:19
  • 1
    $\begingroup$ @sweeneyrod Not sure on this one but,with the binomial theorem you can expand both $a_n=(1+\frac{1}{n})^n $ and $b_n=(1+\frac{x}{n})^{xn}$ and compare them. $\endgroup$ Jun 28, 2015 at 12:11
  • $\begingroup$ @MathematicianByMistake Why is $b_n = (1+ \frac{x}{n})^{xn}$ if the modified form is $(1+\frac{x}{n})^n$. $\endgroup$
    – user716881
    May 30, 2021 at 18:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .