$\dfrac{d}{dx}{\ln{x}}$ formula Proof
Fact-checked:
Formula
$\dfrac{d}{dx}{\ln{x}}$ $\,=\,$ $\dfrac{1}{x}$
$x$ is a variable and the natural logarithm of $x$ is written as $\log_{e}{x}$ or $\ln{x}$ in logarithmic mathematics. Now, differentiate logarithm of $x$ with respect to $x$.
The differentiation of logarithm of $x$ with respect to $x$ is written as $\dfrac{d}{dx}\log_{e}{x}$ (or) $\dfrac{d}{dx}\ln{x}$ in calculus.
Express derivative of log function in limit form
According to derivative of a function with respect to $x$ in the limit form,
$\dfrac{d}{dx} \, f{(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{f{(x+h)}-f{(x)}}{h}$
Take $f{(x)} = \log_{e}{x}$, then $f{(x+h)} = \log_{e}{(x+h)}$. Now, express derivative of log function in limit form.
$\dfrac{d}{dx} \log_{e}{x}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\log_{e}{(x+h)}-\log_{e}{x}}{h}$
Use Quotient rule of logarithms
Try quotient law of logarithms to combine the difference of the logarithmic functions.
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\log_{e}{\Bigg(\dfrac{x+h}{x}\Bigg)}}{h}$
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\log_{e}{\Bigg(\dfrac{x}{x}+\dfrac{h}{x}\Bigg)}}{h}$
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\log_{e}{\Bigg(\require{cancel} \dfrac{\cancel{x}}{\cancel{x}}+\dfrac{h}{x}\Bigg)}}{h}$
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{h}{x}\Bigg)}}{h}$
Expand the logarithmic function
According to logarithmic mathematics, the $\ln{(1+x)}$ can be expanded as an infinite series.
$\log_{e}{(1+x)}$ $\,=\,$ $x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\ldots$
In our case, $\log_{e}{\Bigg(1+\dfrac{h}{x}\Bigg)}$ is a logarithmic function. It can be expanded in the same way by replacing $x$ by $\dfrac{h}{x}$.
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{ \dfrac{h}{x}-\dfrac{{\Bigg(\dfrac{h}{x}\Bigg)}^2}{2}+\dfrac{{\Bigg(\dfrac{h}{x}\Bigg)}^3}{3}-\dfrac{{\Bigg(\dfrac{h}{x}\Bigg)}^4}{4}+\ldots}{h}$
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\dfrac{h}{x}-\dfrac{h^2}{2x^2}+\dfrac{h^3}{3x^3}-\dfrac{h^4}{4x^4}+\ldots}{h}$
$\dfrac{h}{x}$ is a common factor in each term of the infinite series. Take it common from all the terms in the numerator and then simplify the whole function.
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{\dfrac{h}{x}\Bigg[1-\dfrac{h}{2x}+\dfrac{h^2}{3x^2}-\dfrac{h^3}{4x^3}+\ldots\Bigg]}{h}$
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{h\Bigg[1-\dfrac{h}{2x}+\dfrac{h^2}{3x^2}-\dfrac{h^3}{4x^3}+\ldots\Bigg]}{hx}$
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \require{cancel} \dfrac{\cancel{h}\Bigg[1-\dfrac{h}{2x}+\dfrac{h^2}{3x^2}-\dfrac{h^3}{4x^3}+\ldots\Bigg]}{\cancel{h}x}$
$= \,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{1-\dfrac{h}{2x}+\dfrac{h^2}{3x^2}-\dfrac{h^3}{4x^3}+\ldots}{x}$
Evaluate the function
Finally, find the value of the function as the limit $h$ approaches zero. It can be done by replacing $h$ by zero.
$= \,\,\,$ $\dfrac{1-\dfrac{(0)}{2x}+\dfrac{{(0)}^2}{3x^2}-\dfrac{{(0)}^3}{4x^3}+\ldots}{x}$
$= \,\,\,$ $\dfrac{1-0+0-0+\ldots}{x}$
$= \,\,\,$ $\dfrac{1}{x}$
Therefore, it is proved in calculus that the derivative of $\ln{x}$ with respect to $x$ is equal to $\dfrac{1}{x}$
$\,\,\, \therefore \,\,\,\,\,\, \dfrac{d}{dx}{\ln{x}}$ $\,=\,$ $\dfrac{1}{x}$
