A version of the product rule

Question

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

$$\frac{ {\rm d} K(f,g)}{ {\rm d} x} = \frac{f'g-fg'}{fg}$$

...?

Answer 1

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.

Answer 2

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

• products into sums: $\;\dfrac{(fg)'}{fg}=\dfrac{f'}{f}+\dfrac{g'}{g}$,
• quotients into differences: $\;\dfrac{(f/g)'}{f/g}=\dfrac{f'}{f}-\dfrac{g'}{g}$,
• exponents into factors: $\dfrac{\bigl(f^\alpha\bigr)'}{f^\alpha}=\alpha.\dfrac{f'}{f}$

It can make computing derivatives with many factors/ exponents easier:

Example: $\;f(x)=\mathrm e^{x^2}\sqrt{\dfrac{x-1}{x+1}}\dfrac{\sqrt[3]{x^2+1}}{\sqrt[5]{x-2}}$

$$\frac{f'(x)}{f(x)}=2x+\frac12\Bigl(\frac1{x-1}-\frac1{x+1}\Bigr)+\frac13\frac{2x}{x^2+1}-\frac15\frac1{x-2}=\text{&c.}$$