Question

f’(pi/3) for f(x) = ln (cos(x)) ?

Answer 1

`-sqrt(3)`

Explanation:

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)`

Answer 2

`-sqrt(3)`

Explanation:

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

Answer 3

If `f(x) = ln (cos(x))`, then `f’(pi/3)= -sqrt(3)`

Explanation:

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)`