How $\ln{f} = a_{1}\ln{x_{1}} + a_{2}\ln{x_{2}} +...+ a_{k}\ln{x_{k}}$ is differentiated here?

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I'm learning about how to evaluate error in measurement and I don't understand how logarithmic derivative method (which is the only method I know about) is derivated. It says:

When the function $f$ is the product of variables $x_{1}...x_{k}$, the log derivative method is used to find the uncertainty of $f$. $$ f = x_{1}^{a_{1}}\cdot \hspace{1mm}...\hspace{1mm}\cdot x_{k}^{a_{k}} $$

Take the natural log of each side: $$ \ln{f} = a_{1}\ln{x_{1}} + a_{2}\ln{x_{2}} +...+ a_{k}\ln{x_{k}} $$ Then differentiate: $$ \displaystyle \frac{\Delta f}{f} = a_{1}\frac{\Delta x_{1}}{x_{1}} +\hspace{1mm}... \hspace{1mm} + a_{k} \frac{\Delta x_{k}}{x_{k}} $$

Here $\Delta f$ is the absolute error. So what I know here is that the derivative of $\ln{(f(x))}$ is $\displaystyle \frac{f'(x)}{f(x)}$ but this is not what I see in the differentiation, also is it some kind of total differentiation? because it isn't done with respect to one variable.

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Consider all the $x_i$ to be functions of some parameter $t$. Since $f$ is a function of the $x_i$, this also makes $f$ a function of $t$. So the derivative with respect to $t$ is

$$\frac{d}{dt}\ln(f) = a_1\frac{d}{dt}\ln x_1 + \dots + a_k\frac{d}{dt}\ln x_k$$ $$\frac 1{f}\frac{df}{dt} = a_1\frac 1{x_1}\frac{dx_1}{dt} + \dots + a_k\frac 1{x_k}\frac{dx_k}{dt}$$ Multiplying through by $dt$, $$\frac{df}{f} = a_1\frac{dx_1}{x_1} + ... + a_k\frac{dx_k}{x_k}$$

When finite differences $\Delta f, \Delta x_i$ are very small, they behave nearly like infinitesimals. The equation you are after is making this approximation. So replace each of the $df, dx_i$ with $\Delta f, \Delta x_i$, and you get

$$\frac{\Delta f}{f} = a_1\frac{\Delta x_1}{x_1} + ... + a_k\frac{\Delta x_k}{x_k}$$