Logarithmus

$$\log \left(\prod _{i=1}^n{y}_i\right)=\sum _{i=1}^n{y}_i$$

Ableiten vom Logarithmus#

$$\frac{d}{db}a\cdot log(b)=\frac{a}{b}$$ $$\frac{d}{da}log(a)=\frac{1}{a}$$ $$\frac{d}{da}log(1-a)=\frac{-1}{1-a}$$