Proof of the derivative of $\ln(x)$

Question

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?

Answer 1

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:

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

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

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

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

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

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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.

Answer 6

An alternative method would start with: -

$$y = \ln(x)\hspace{1cm}\rightarrow\hspace{1cm} e^y = x$$

Basic differentiation from 1st principles: -

$$e^{y+\delta y} = x+\delta x\hspace{1cm}\rightarrow\hspace{1cm} e^y\cdot e^{\delta y} = x+\delta x$$

It follows that: -

$$e^{\delta y} = 1+\dfrac{\delta x}{x}$$

Using the series expansion of $e^y$ and reducing $\delta y$ to zero we get: -

$$1+dy = 1+\dfrac{dx}{x}$$

Hence, $\dfrac{dy}{dx} = \dfrac{1}{x}$