f’(pi/3) for f(x) = ln (cos(x)) ?

First you need to find #f'(x)#

hence, #(df(x))/dx = (d[ln(cos(x))])/dx#

we will apply chain rule in here,

so #(d[ln(cos(x))])/dx = 1/cos(x) *(-sinx)#.........................(1)

since, #(d[ln(x)]/dx = 1/x and d(cos(x))/dx = -sinx)#

and we know #sin(x)/cos(x)= tanx#

hence the above equation(1) will be

                       # f'(x)=-tan(x) #

and, #f'(pi/3)=-(sqrt3)#

#f(x)=ln(cos(x))#
#f'(x)=-sin(x)/cos(x)=-tan(x)#
#f'(pi/3)=-tan(pi/3)=-sqrt(3)#

The expression #ln(cos(x))# is an example of function composition.

Function composition is in essence just combining two or more functions in a chain to form a new function -- a composite function.

When evaluating a composite function, the output of an inner component function is used as the input to the outer likes links in a chain.

Some notation for composite functions: if #u# and #v# are functions, the composite function #u(v(x))# is often written #u circ v# which is pronounced "u circle v" or "u following v."

There is a rule for evaluating the derivative of these functions composed from chains of other functions: the Chain Rule.

The Chain Rule states:

#(u circ v)'(x) = u'(v(x)) * v'(x)#

The Chain Rule is derived from the definition of derivative.

Let #u(x) = ln x#, and #v(x) = cos x#. This means that our original function #f = ln(cos(x)) = u circ v#.

We know that #u'(x) = 1/x# and #v'(x) = -sin x#

Restating the Chain Rule and applying it to our problem:

#f'(x) = (u circ v)'(x)#

#\ \ \ \ \ \ = u'(v(x)) * v'(x)#

#\ \ \ \ \ \ = u'(cos(x)) * v'(x)#

#\ \ \ \ \ \ = 1/cos(x) * -sin(x)#

#\ \ \ \ \ \ = -sin(x)/cos(x)#

#\ \ \ \ \ \ = -tan(x)#

It is a given that #x = pi/3#; therefore,

#f’(pi/3)= -tan(pi/3) = -sqrt(3)#