How do you differentiate $\left( {\cos x} \right)\left( {\sin x} \right)$?

Hint: We will first divide and multiply the given expression by 2, then we will use the fact that $\sin 2\theta = 2\sin \theta \cos \theta $. Thus, we have simplified expressions which can be easily differentiated.

Complete step by step solution:
We are given that we are required to differentiate $\left( {\cos x} \right)\left( {\sin x} \right)$.
Let us assume that $f(x) = \left( {\cos x} \right)\left( {\sin x} \right)$.
Now, multiplying and dividing the given function by 2, we will then obtain the following equation with us:-
$ \Rightarrow f(x) = \dfrac{{2\sin x\cos x}}{2}$
Now, since we know that we have a formula given by $\sin 2\theta = 2\sin \theta \cos \theta $ for any $\theta \in \mathbb{R}$.
Using this, we will obtain the following equation with us:-
$ \Rightarrow f(x) = \dfrac{{\sin 2x}}{2}$
Now, differentiating both the sides of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow \dfrac{d}{{dx}}\left\{ {f(x)} \right\} = \dfrac{d}{{dx}}\left( {\dfrac{{\sin 2x}}{2}} \right)$
Simplifying the above equation, we will then obtain the following equation with us:-
$ \Rightarrow f'(x) = \dfrac{1}{2}\cos 2x \times 2$
Simplifying the right hand side of the above equation further, we will then obtain the following equation with us:-

$ \Rightarrow f'(x) = \cos 2x$

Thus, we have the required answer with us.

Note:
The students must note that we have used the fact that the differentiation of sine of any angle is given by cosine of it which can be written in the form of equation as follows:-
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sin \theta } \right) = \cos \theta $
Now, we have just added up the chain rule in this fact as well which states that, if we have f (g (x)), then its differentiation is given by the following expression:-
$ \Rightarrow \dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)$
Here, just replace f (x) by sin x and g (x) by 2x ad thus, we have the equation required as done in the above solution.

Alternate Way:
We can use the chain rule of differentiation as well which states that if we have two functions u(x) and v(x), then the differentiation of their product is given by the following expression:-
$ \Rightarrow \dfrac{d}{{dx}}\left\{ {u\left( x \right).v\left( x \right)} \right\} = v\left( x \right)\dfrac{d}{{dx}}\left\{ {u\left( x \right)} \right\} + u\left( x \right)\dfrac{d}{{dx}}\left\{ {v\left( x \right)} \right\}$
Replacing u(x) by sin x and v(x) by cos x, thus we get:-
$ \Rightarrow \dfrac{d}{{dx}}\left\{ {\sin x.\cos x} \right\} = \cos x\dfrac{d}{{dx}}\left\{ {\sin x} \right\} + \sin x\dfrac{d}{{dx}}\left\{ {\cos x} \right\}$
$ \Rightarrow \dfrac{d}{{dx}}\left\{ {\sin x.\cos x} \right\} = {\cos ^2}x - {\sin ^2}x$
$ \Rightarrow \dfrac{d}{{dx}}\left\{ {\sin x.\cos x} \right\} = \cos 2x$