What is the derivative of $\log \left( \cos x \right)$?

Hint: We solve this problem by using the chain rule of differentiation and some standard formulas of differentiation.
(i) The chain rule of differentiation is given as
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$
(ii) The derivative of cosine function is given as,
$\dfrac{d}{dx}\left( \cos x \right)=-\sin x$
(iii) The derivative of logarithm function is given as,
$\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$
By using the above three formulas we can find the required value.

Complete step by step solution:
Let us assume the given function as,
$\Rightarrow y=\log \left( \cos x \right)$
Now, let us assume two functions $f\left( x \right),g\left( x \right)$ as,
$\begin{align}
  & \Rightarrow f\left( x \right)=\log x \\
 & \Rightarrow g\left( x \right)=\cos x \\
\end{align}$
Here, we can reframe the given equation as,
$\Rightarrow y=f\left( g\left( x \right) \right)$
Now, let us differentiate both sides with respect to $'x'$ then we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)$
We know that the chain rule of differentiation is given as
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$
By using the chain rule to above reframed equation then we get,
\[\Rightarrow \dfrac{dy}{dx}={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)............equation(i)\]
Now, let us take the first function $f\left( x \right)$ and by differentiating it with respect to $'x'$ then we get,
\[\Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( \log x \right)\]
We know that the derivative of logarithm function is given as,
$\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$
By using this rule in above differentiation then we get,
$\Rightarrow {f}'\left( x \right)=\dfrac{1}{x}$
Now, by replacing $'x'$ by $\left( \cos x \right)$ in above equation then we get,
$\begin{align}
  & \Rightarrow {f}'\left( g\left( x \right) \right)=\dfrac{1}{\left( \cos x \right)} \\
 & \Rightarrow {f}'\left( g\left( x \right) \right)=\dfrac{1}{\left( \cos x \right)} \\
\end{align}$
Now, let us take the second function that is $g\left( x \right)$ and by differentiating it with respect to $'x'$ then we get,
$\Rightarrow {g}'\left( x \right)=\dfrac{d}{dx}\left( \cos x \right)$
We know that the derivative of cosine function is given as,
$\dfrac{d}{dx}\left( \cos x \right)=-\sin x$

By using the above formula in the ${g}'\left( x \right)$ then we get,
$\Rightarrow {g}'\left( x \right)=-\sin x$
Now, let us take equation (i) and substitute the required values then we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\left[ \dfrac{1}{\left( \cos x \right)} \right]\times \left[ -\sin x \right] \\
 & \Rightarrow \dfrac{dy}{dx}={-\tan x} \\
\end{align}$

Therefore, the derivative of given equation can be written as,
$\therefore \dfrac{dy}{dx}={-\tan x}$

Note: The common mistake that can be done in this question of taking the formula of the differentiation. The chain rule is applied when the given equation includes one function inside the second equation and it is given as,
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$
But students may do mistake without taking the differentiation of second function and take the formula as,
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={f}'\left( g\left( x \right) \right)$
This gives the wrong answer. So, we need to take care of formulas to get the required answer correctly.
Also the derivative of logarithm function should be taken as,
$\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$