ne amintim ca f(g(x)) '= f'(g(x)) *g'(x)
si ca (fg)'=f'g+fg'
si ca (f/g)' = (f'g-fg')/g²
asadar
1) f'= (3^2x)ln3*(2x)'-[1/(2x-1)]*(2x-1)'
adica
2*3^(2x)ln3 -2/(2x-1)
2) asta e mai greu din cauza lui lgx
il rezolv asa cum e dat
(x²)' lgx +x²* (lgx)'
(x²)'=2x
lgx=lnx*lge, in care lge este o constanta, deci la derivare , iese in fata
(lgx)'= (lnx*lge)'=lge*(lnx)"=lge*1/x
asadar f'=2xlgx+(x²) (lge)/x=2x *(lgx) +x² lge (1/x)=
=2xlgx +xlge=
=x(2lgx+lge)
3)f'= [(lnx)' *x³-(lnx)* (x³)']/((x³)²)=
[1/x)*x³- (3x²)*lnx]/x^6=
(x² -3x²lnx)/x^6=
x²(1-3lnx)/x^6=
(1-3lnx)/x^4
pt D' ⊆D
