Find f'(x) if f(x)=logbasex(x^2-5x+6).
2007-02-20 23:46:56 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
f(x) = log basex (x^2-5x+6)
f(x) = ln (x^2-5x+6) / ln x
f'(x) = (2x-5) / [(x^2 - 5x + 6) ln x] - ln (x^2-5x+6) / [x (ln x)^2]
use another method
f(x) = log basex (x^2-5x+6)
x^f(x) = x^2-5x+6
ln x^f(x) = ln (x^2-5x+6)
f(x) ln x = ln (x^2-5x+6)
differentiate both side
f'(x) ln x + f(x) (1/x) = (2x - 5) / (x^2-5x+6)
f'(x) ln x + log basex (x^2-5x+6) (1/x) = (2x - 5) / (x^2-5x+6)
f'(x) ln x + (ln(x^2-5x+6)/ln x) (1/x) = (2x - 5) / (x^2-5x+6)
Divide both side by ln x
f'(x) + (ln(x^2-5x+6)/(ln x)^2) (1/x) = (2x - 5) / [(x^2-5x+6) ln x]
f'(x) = (2x-5) / [(x^2 - 5x + 6) ln x] - ln (x^2-5x+6) / [x (ln x)^2]
Both method get same answer!!!
2007-02-20 23:57:59 · answer #1 · answered by seah 7 · 1⤊ 0⤋
y=LN[ (x) *(x^2-5x+6) ]
y'=[1/(x * (x^2-5x+6))] *[ (x^2-5x+6)*1 + x *(2x-5) ]
2007-02-20 23:56:57 · answer #2 · answered by fcas80 7 · 0⤊ 0⤋