Using the differentiation rules and the derivative table, find the derivative of the function y=\frac{1}{\ln(\cos(x))}


y'=\frac{\tan x}{\ln^2(\cos x)}

We can read the function as composed in this way.

  • Identify the outer function and inner function

    y = \frac{1}{g(x)}\qquad \text{con}\qquad g(x)=\ln(\cos(x))


    where the outer function is the reciprocal: f(\cdot)=\frac{1}{(\cdot)}. Let’s apply the Chain Rule remembering that the derivative with respect to t of f(t)=\frac1t is f'(t)=-\frac{1}{t^2}:

    y'=-\frac{1}{[g(x)]^2}\cdot g'(x)


    We therefore need g'

  • Calculate the derivative of g

    It is still a composition:

    g(x)=\ln (v(x))\qquad \text{con}\qquad v(x)=\cos(x) 


    Let’s apply the Chain Rule, remembering that v'(x)=-\sin x:

    g'(x)=\frac{1}{v(x)}\cdot v'(x)=\frac{1}{\cos x}\cdot -\sin x=-\tan x


  • Replace g and g' in the Chain Rule


    • y'=-\frac{1}{[\ln(\cos x)]^2}\cdot -\tan x


    Simplifying, the final result is

    y'=\frac{\tan x}{[\ln(\cos x)]^2}