Question

derivative of log(sin^2(x))

Answer 1

2cotx

Answer 2

Thank you, but how do you get that?

Answer 3

Chlorophyll's method.

Answer 4

Got it, thanks guys :)

Answer 5

using chain rule
let u=sin^2(x)
du/dx=2sin(x)cos(x)
thus, y=log(u)
dy/du=1/(u*ln(u)) = 1/(sin^2(x)*ln(sin^2(x))
y'=du/dx * dy/du = 2sin(x)cos(x) * 1/(sin^2(x)*ln(sin^2(x))
simplify it !

Answer 6

i think ash.vasvani's and chlorophyll's answer still wrong
because derivative of log(u) is not u'/u but should be 1/(u'*ln(u))

Answer 7

am i right, @Chlorophyll ?

Answer 8

I'm sorry to say you should double check the formula :)

Answer 9

ok, nope... sometimes i do mistake too :D

Answer 10

Yes, you do make mistake this time :)
As most people know (ln x)' = x'/x = 1/x
similarly, (ln u)' = u'/ u

Answer 11

but, for this time u have mistaken again :P
the original problem is y=log(...) not y=ln(...)
hahahaha...

Answer 12

I see what you mean now!
Yes, I read the post so fast, however double check your the formula, still:
y = log ( u ) --> y' = u' / u ln a

Answer 13

yups, if like that i with u :)
but a=base10, right ?

Answer 14

Yep, to make it clear
y = log ( u ) --> y' = u' / (u ln10)

Answer 15

ok, thank u...
nice discuss with u brother :)

Answer 16

It would be nice to go to the bottom of discussion :)

Answer 17

yea, but for this time i will go, i have a class know
my student will angry to me if come late :)

Answer 18

nothink, bye...