Дана функция [m] \varphi(x) = \frac{1}{6} \ln (-3x) [/m]. Найдите [m] \varphi'(x) [/m], [m] \varphi' \left( -\frac{1}{9} \right) [/m].
φ `(x)=(1/6)·(1/(–3x))·(–3x)`=(1/6)·(1/(–3x))·(–3)=1/(6x) φ `(–1/9)=1/(–6/9)=–3/2