Using the product rule we know that
$$\frac{ {\rm d}\ln(fg)}{ {\rm d} x} = \frac{f'g+fg'}{fg}$$
Is there a function $K$ such that
$$\frac{ {\rm d} K(f,g)}{ {\rm d} x} = \frac{f'g-fg'}{fg}$$
...?
Using the product rule we know that
$$\frac{ {\rm d}\ln(fg)}{ {\rm d} x} = \frac{f'g+fg'}{fg}$$
Is there a function $K$ such that
$$\frac{ {\rm d} K(f,g)}{ {\rm d} x} = \frac{f'g-fg'}{fg}$$
...?
Note that the first formula works due to $$ \log (fg) = \log f + \log g $$ All we have to do is changing the sign in front of $\log g $, hence $$ K ( f, g) = \log f - \log g = \log\frac fg $$ will do the trick.
A comment or an extension for this question: $\dfrac{\mathrm d\mkern1mu(\ln\lvert f\rvert)}{\mathrm d\mkern1mu x}$ is known as the logarithmic derivative of $f$, and it inherits the properties of the logarithmic function: it turns
It can make computing derivatives with many factors/ exponents easier:
Example: $\;f(x)=\mathrm e^{x^2}\sqrt{\dfrac{x-1}{x+1}}\dfrac{\sqrt[3]{x^2+1}}{\sqrt[5]{x-2}}$
$$\frac{f'(x)}{f(x)}=2x+\frac12\Bigl(\frac1{x-1}-\frac1{x+1}\Bigr)+\frac13\frac{2x}{x^2+1}-\frac15\frac1{x-2}=\text{&c.}$$