f’(pi/3) for f(x) = ln (cos(x)) ?

3 Answers
May 26, 2018

-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)

May 26, 2018

-sqrt(3)

Explanation:

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

May 26, 2018

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)