Why does logarithmic differentiation work even if logs are not defined for negative numbers?

Question

jean2098 · Accepted Answer

Logarithms are defined for all numbers except 0.
If you haven’t learned about Euler’s identity, it states that

e^{π i} = - 1

This is a very famous identity, which is learnt typically during the early stages of complex numbers.
Anyway, this formula basically shows that we can take a

\log

of a negative number, we’ll just get a complex result.

\log (e^{π i}) = \log (- 1)

Hence,

π i \log (e) = \log (- 1)

and

\log (- 1) = π i

Using the logarithmic identities like the addition one we can then determine the logarithm for any real number except for 0 (you can find out why with some pen and paper).
Since the calculus works for complex numbers as well, albeit differently, you can, in fact, differentiate it.

hysgubwyri3 · Accepted Answer

It makes sense to consider the quantity

\frac{f' (x)}{f (x)}

even without knowing about logarithms. Assume we have a function

x \to y = f (x)

, and consider a working point

(x_{0}, y_{0})

. When studying increments

△ x, △ y

measured from this point the absolute change of y is approximated by

△ y \approx f' (x_{0}) △ x (| △ x | l l 1),

and the relative change of y is approximated by

\frac{△ y}{y_{0}} = \frac{f' (x_{0})}{f (x_{0})} △ x (| △ x | l l 1) .

RizerMix · Accepted Answer

Either using the chain rule and ddx|x|=x|x|, or working cases for x&amp;gt;0 and x&amp;lt;0, we getddxlog⁡(|x|)=1xThus, we getf′(x)f(x)=ddxlog⁡(|f(x)|)

Why does logarithmic differentiation work even if logs are not

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