given y find y'?

2 ⤊

1) y=logx-2logx

2007-11-21 22:09:29 · 5 answers · asked by Jackline W 1 in Science & Mathematics ➔ Mathematics

y = log x - 2 log x
dy/dx = 1 / x - 2 / x = - 1 / x

2007-11-22 01:27:50 · answer #1 · answered by Como 7 · 1⤊ 0⤋

y'= 1/x -2/x = -1/x

2007-11-21 22:16:13 · answer #2 · answered by mbdwy 5 · 0⤊ 0⤋

here: y' = (1/x) - 2(1/x)

using the formula d(logx)/dx= (1/x)
:) hope that helped you..

2007-11-21 22:14:01 · answer #3 · answered by Muslim Girl 2 · 0⤊ 0⤋

y=logx-2logx
y= - log X
y' = - 1/x

2007-11-21 22:25:17 · answer #4 · answered by pioneers 5 · 0⤊ 1⤋

y=logx-2logx
y = - log x . . . . . . . no further simplification

2007-11-21 22:30:06 · answer #5 · answered by CPUcate 6 · 1⤊ 0⤋