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?

  • 15
    $\begingroup$ 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 $\endgroup$ Sep 28, 2021 at 12:33

5 Answers 5


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:

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

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

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

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

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

  • 12
    $\begingroup$ 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
  • 29
    $\begingroup$ 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

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 $$(\ln)'(y)=\frac1{\mathrm e^{\,\ln y}}=\frac1y.$$


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.


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

  • 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

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.

  • 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

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.