How do you differentiate $\operatorname{f}\left( x \right)={{\cos }^{2}}x$ ?

Hint: Problems on differentiating a function like this can be done by simply applying the laws of differentiation accordingly followed by some simplifications. We differentiate the given composite function using the chain rule of differentiation i.e., $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={f}'\left( g\left( x \right) \right){g}'\left( x \right)$ . We assume the function $\cos x$ to be similar to $\operatorname{g}\left( x \right)$ and apply the chain rule which leads us to the required given.

Complete step by step answer:
The function we are given is $\operatorname{f}\left( x \right)={{\cos }^{2}}x$
As the above function is a composite one, we must use the chain rule of differentiation to find the derivative of this.
According to the chain rule of differentiation $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={f}'\left( g\left( x \right) \right){g}'\left( x \right)$
Similarly, we assume the function $\cos x$ to be similar to $\operatorname{g}\left( x \right)$ i.e., $\operatorname{g}\left( x \right)=\cos x$ .
We know
$\Rightarrow \operatorname{f}\left( x \right)={{\cos }^{2}}x={{\left( \cos x \right)}^{2}}$
Hence, from the above assumptions we can write
$\Rightarrow \operatorname{f}\left( \operatorname{g}\left( x \right) \right)={{\left( \cos x \right)}^{2}}={{\left\{ \operatorname{g}\left( x \right) \right\}}^{2}}$
Also, we know
$\Rightarrow {g}'\left( x \right)=\dfrac{d\left\{ \operatorname{g}\left( x \right) \right\}}{dx}$
Completing the above derivation, we get
$\Rightarrow {g}'\left( x \right)=\dfrac{d}{dx}\left( \cos x \right)=-\sin x$
Therefore, applying the chain rule of differentiation we can write
$\Rightarrow \dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=\dfrac{d\operatorname{f}\left( \operatorname{g}\left( x \right) \right)}{d\operatorname{g}\left( x \right)}\cdot \dfrac{d\operatorname{g}\left( x \right)}{dx}$
Also,
$\Rightarrow \dfrac{d\operatorname{f}\left( \operatorname{g}\left( x \right) \right)}{d\operatorname{g}\left( x \right)}=\dfrac{d{{\left\{ \operatorname{g}\left( x \right) \right\}}^{2}}}{d\operatorname{g}\left( x \right)}$
Completing the derivation, we get
$\Rightarrow \dfrac{d\operatorname{f}\left( \operatorname{g}\left( x \right) \right)}{d\operatorname{g}\left( x \right)}=2\cos x$
Now, we substitute the above values of differentiation of the functions in the main equation as shown below
$\Rightarrow \dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=\dfrac{d\operatorname{f}\left( \operatorname{g}\left( x \right) \right)}{d\operatorname{g}\left( x \right)}\cdot \dfrac{d\operatorname{g}\left( x \right)}{dx}$
$\Rightarrow \dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=2\cos x\left( -\sin x \right)$
Omitting the bracket, we get
$\Rightarrow \dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=-2\cos x\sin x$
We know that from the principles of trigonometric functions the formula for double angle identity of sin is
$\sin 2x=2\sin x\cos x$
Hence, we put the above shown expression in place of $2\sin x\cos x$ as shown below
$\Rightarrow \dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=-\sin 2x$
Thus,
$\Rightarrow {f}'\left( x \right)=-\sin 2x$

Therefore, we conclude that the derivative of the given function is $-\sin 2x$

Note: While applying the chain rule of differentiation we must be careful that all the functions are defined properly and the chain rule is used accordingly. Also, in spite of using the chain rule we could have also differentiated the given function by using the first principle of derivative i.e., the conventional method. Though, we prefer using the chain rule as it is simple and easy to understand.