The derivative of logarithmic function can be derived in differential calculus from first principle. $f{(x)}$ is a function in terms of $x$ and the natural logarithm of the function $f{(x)}$ is written as $\log_{e}{f(x)}$ or $\ln{f{(x)}}$ in mathematics. The differentiation of logarithmic function with respect to $x$ is written in mathematical form as follows.
$\dfrac{d}{dx}{\, \log_{e}{f{(x)}}}$
According to the definition of the derivative, the differentiation of a function with respect to a variable can be written in limiting operation form.
$\dfrac{d}{dx}{\, g{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{g{(x+\Delta x)}-g{(x)}}{\Delta x}}$
In this case, $g{(x)} = \log_{e}{f{(x)}}$ then $g{(x+\Delta x)} = \log_{e}{f{(x+\Delta x)}}$. Now, write the derivative of the natural logarithm of a function in the form of limiting operation.
$\implies$ $\dfrac{d}{dx}{\, \log_{e}{f{(x)}}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{f{(x+\Delta x)}}-\log_{e}{f{(x)}}}{\Delta x}}$
According to differential, take $f{(x)} = y$ and $f{(x+\Delta x)} = y+\Delta y$. Now, write each function in terms of $y$.
$\implies$ $\dfrac{d}{dx}{\, \log_{e}{f{(x)}}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{(y+\Delta y)}-\log_{e}{y}}{\Delta x}}$
The difference of the logarithmic terms can be combined by using quotient rule of logarithms.
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{y+\Delta y}{y}\Bigg)}}{\Delta x}}$
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{y}{y}+\dfrac{\Delta y}{y}\Bigg)}}{\Delta x}}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(\dfrac{\cancel{y}}{\cancel{y}}+\dfrac{\Delta y}{y}\Bigg)}}{\Delta x}}$
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\Delta x}}$
The limiting operation can be evaluated by using limit rule of logarithmic function. So, try to transform it same as the standard result.
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\Delta x \times 1}}$
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\Delta x \times \dfrac{\dfrac{\Delta y}{y}}{\dfrac{\Delta y}{y}} }}$
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\Delta x \times \dfrac{\Delta y}{y} \times \dfrac{y}{\Delta y} }}$
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\dfrac{\Delta y}{y} \times \Delta x \times \dfrac{y}{\Delta y} }}$
Now, factorize the function in the limiting operation.
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \Bigg[\dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\dfrac{\Delta y}{y}}}$ $\times$ $\dfrac{1}{\Delta x \times \dfrac{y}{\Delta y}} \Bigg]$
$=\,\,\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \Bigg[\dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\dfrac{\Delta y}{y}}}$ $\times$ $\dfrac{\Delta y}{\Delta x \times y} \Bigg]$
$\implies$ $\dfrac{d}{dx}{\, \log_{e}{f{(x)}}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \Bigg[\dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\dfrac{\Delta y}{y}}}$ $\times$ $\dfrac{\Delta y}{\Delta x}$ $\times$ $\dfrac{1}{y}\Bigg]$
Use the product rule of limits to find the limit of the function by the product of their limits.
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\dfrac{\Delta y}{y}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\Delta y}{\Delta x}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{1}{y}\Bigg)}$
According to concept of derivative, if the change in $x$ is infinitesimal then the change in $y$ is also infinitesimal. Therefore, $\Delta y \to 0$, if $\Delta x \to 0$.
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{\Delta y \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\dfrac{\Delta y}{y}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\Delta y}{\Delta x}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{1}{y}\Bigg)}$
If $\Delta y \to 0$, then $\dfrac{\Delta y}{y} \to \dfrac{0}{y}$. Therefore, $\dfrac{\Delta y}{y} \to 0$
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{\frac{\Delta y}{y} \,\to\, 0}{\normalsize \dfrac{\log_{e}{\Bigg(1+\dfrac{\Delta y}{y}\Bigg)}}{\dfrac{\Delta y}{y}}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\Delta y}{\Delta x}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{1}{y}\Bigg)}$
Take $m = \dfrac{\Delta y}{y}$ and change the first factor in terms of $m$.
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\log_{e}{(1+m)}}{m}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\Delta y}{\Delta x}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{1}{y}\Bigg)}$
According to logarithmic limit rule, the limit of the first logarithmic function is one.
$=\,\,\,$ $1$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\Delta y}{\Delta x}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{1}{y}\Bigg)}$
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\Delta y}{\Delta x}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{1}{y}\Bigg)}$
Actually, $y = f{(x)}$ and $y+\Delta y = f{(x+\Delta x)}$. Therefore, $\Delta y = f{(x+\Delta x)}-f{(x)}$. Replace them in the both factors.
$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f{(x+\Delta x)}-f{(x)}}{\Delta x}\Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{1}{f{(x)}}\Bigg)}$
According to concept of derivative, the value of the first factor is equal to the derivative of the function $f{(x)}$ with respect to $x$. Now, evaluate the third factor by the direct substitution method. The limit of reciprocal of $f{(x)}$ as change in $x$ approaches zero is remain same due to lack of change in $x$ term in the function.
$=\,\,\,$ $\Bigg(\dfrac{d}{dx}{\, f{(x)}}\Bigg)$ $\times$ $\Bigg(\dfrac{1}{f{(x)}}\Bigg)$
$=\,\,\,$ $\Bigg(\dfrac{1}{f{(x)}}\Bigg)$ $\times$ $\Bigg(\dfrac{d}{dx}{\, f{(x)}}\Bigg)$
$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \log_{e}{f{(x)}}}$ $\,=\,$ ${\dfrac{1}{f{(x)}}}{\dfrac{d}{dx}{\, f{(x)}}}$
Therefore, it is proved that the derivative of logarithm of a function with respect to variable is equal to the product of the reciprocal of the function and the derivative of the function.
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