# Proof of the derivative of ln(x)

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 $e^{\frac{1}{x}} = \lim_{h\to0}\left (1 + h \right )^{\frac{1}{xh}}$, 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?

• Which definition of the natural logarithm do you have? – Bernard Jun 28 '15 at 10:29
• $\ln$ 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? – krvolok Jun 28 '15 at 10:30
• @Bernard I have the natural logarithm defined as the inverse of $e^{x}$. – rlms Jun 28 '15 at 10:35
• @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. – rlms Jun 28 '15 at 10:37

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.

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:

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

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

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

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

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

• 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. – Chaitanya Tappu Jun 28 '15 at 12:07
• @cmtappu96 Indeed, good point. – wythagoras Jun 28 '15 at 12:13
• The inverse function theorem guarantees that $\ln x$ is differentiable. – mweiss Jun 28 '15 at 16:33

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

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

• For this method, you also need to know that $\: \ln \:$ is differentiable. $\;\;\;\;$ – user57159 Jun 28 '15 at 23:10

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.

• How do you get from the definition to the slightly modified form? – rlms Jun 28 '15 at 11:19
• @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. – MathematicianByMistake Jun 28 '15 at 12:11