Răspunsuri

2014-06-02T18:15:25+03:00
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
Facem mai intai fara substitutie, ridicand la puterea a treia avem: (x^3+3x^2+3x+1)`
acum derivam:3x^2+6x+3
Daca facem cu substitutie avem u=x+1 (u^3)`=3u^2*u` 3*(x+1)^2*(x+1)` =3(x^2+2x+1) *(1+0) =3x^2+6x+3
u=x+1...... (u^3)`=3u^2*u`...... 3*(x+1)^2*(x+1)` =3(x^2+2x+1) *(1+0) =3x^2+6x+3 uneori e mai usor sa faci cu u
o....scuze nu am vazut pus de doua ori rezolvarea derivatei, am avut unele greseli de scriere prima data....-_-