Help me with Calculus please?

0 ⤊

Differentiate:
y=lnx(ln2x)

y = lnx times ln2x

2007-04-08 06:33:16 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

answer is:

1/x (lnx+ln2x)

2007-04-08 06:49:41 · update #1

1)
y'= (1 ln (2x) + (2/2x) x )/ (x ln(2x))

2)
y'= (1/x) (ln2x) + (2/2x) (lnx)

2007-04-08 06:44:57 · answer #1 · answered by iyiogrenci 6 · 0⤊ 0⤋

product rule.

y' = (1/x)(ln2x) + (lnx)(1/2x)(2)
simplifies to
ln2x/x + lnx/x

2007-04-08 13:37:44 · answer #2 · answered by ǝɔnɐs ǝɯosǝʍɐ Lazarus'd- DEI 6 · 0⤊ 0⤋

y = lnx(ln2x)
y = ln(x + 2x)
y = ln(3x)

y' = (1/3x)(3)
y' = 3/3x
y' = 1/x

2007-04-08 13:40:47 · answer #3 · answered by Anonymous · 0⤊ 0⤋