if y= ln (ln 2x), then y` = ?

Question

A) x/ (ln2x)
B) 1/ (x(ln 2x))
C) 1/ (ln 2x)
D) 2/(ln 2x)
E) 1/ (ln (ln 2x))

Answer 1

The derivative of lnx = 1/x * chain of x
So for y= ln (ln 2x), we get
y' = 1/ln2x *1/2x * 2
y' = 1/ln2x *1/x
y' = 1/xln2x

[Answer: B]

Answer 2

y'=1/ln(2x) *d/dx(ln2x) =1/x*1/ln(2x) so the answer is B.

Answer 3

y = ln u where u = ln 2x
dy/du = 1/u and du/dx = 1/x
dy/dx = [ 1 / ln (2x) ] [1 / x ]
dy/dx = [ 1 / (x ln 2x) ]
OPTION B

Answer 4

answer is B using chain rule
y=ln(x)
y'=1/x


so
y=ln(ln( 2x)
y'=1/(ln(2x)*1/(2x)*2
y'=1/(ln(2x)*1/x

Answer 5

let u = ln(2x)

y = ln(u)

dy/dx = dy/du* du/dx = 1/u* 1/x

= 1/(x*ln(2x))

Answer 6

c

Answer 7

c