Question

Cum se calculeaza derivata aceasta? f(x)=ln [(1-x/x-1)]' 

Answer 1

$f`(x)=ln( \frac{1-x}{x-1} ) \\ u= \frac{1-x}{x-1} \\ (ln u)`= \frac{1}{u}*u` \\ f`(x)= \frac{x-1}{1-x} *( \frac{1-x}{x-1})` \\ f`(x)=(\frac{x-1}{1-x} )*[(1-x)`*(x-1)-(1-x)(x-1)`(/x-1)^{2}] \\ f`(x)=( \frac{x-1}{1-x})*$$f`(x)=ln( \frac{1-x}{x-1} ) \\ u= \frac{1-x}{x-1} \\ (ln u)`= \frac{1}{u}*u` \\ f`(x)= \frac{x-1}{1-x} *( \frac{1-x}{x-1})` \\ f`(x)=(\frac{x-1}{1-x} )*[(1-x)`(x-1)-(1-x)(x-1)`/(x-1)^{2}] \\ f`(x)=( \frac{x-1}{1-x})*[(0-1)(x-1)-(1-x)(1-0)/(x-1) ^{2} ] \\ f`(x)= (\frac{x-1}{1-x})*[-1(x-1)-(1-x)/(x-1) ^{2} ] \\ f`(x)=( \frac{x-1}{1-x})*(-x+1-1+x)/(x-1)^2 \\ f`(x)=0$