How do you differentiate this logarithmic func involving trig?

Question

diff:

f(x)= cos (ln x)


A step by step brake down would be appreciated. Trig in anything always messes me up.

Answer 1

f ' = - sin( ln x ) * ( 1 / x)

where deriv of ln ( x) = 1 / x ,

you need to use the chain rule....
f ( w) , where w = ln ( x) , then f ' = ( df / dw ) ( dw / dx)

here df / dw = - sin ( w ) , and dw / dx = ( ln x) ' = 1 / x , replace w with ln x, substitute into f ' above, and you are done...

Answer 2

f'(x) = -sin u du/dx
u=ln x
du/dx=1/x
now replace u with its value
f'(x)= -sin (ln x) *1/x

Answer 3

use the chain rule if you know hhow if not just give me a sec

okay LET u=lnx
the chaine rule is dy/dx=dy/du*du/dx
dy/du=-sinu
du/dx=1/x
so dy/dx = -(1/x)Sin(lnx)

Answer 4

f(x)= cos (ln x)
f'(x) = -sin(ln x) d/dx(lnx)
= -sin(ln x) [1/x]
= -[sin(ln x)]/x

Answer 5

(lnx)'=1/x

(cos u)'=-u'sinu

Hence
f ' (x)= - (1/x) sin(1/x)