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

Hint: To solve this problem we will use the concept of chain rule of differentiation. In this rule, we will assume the inner function as $t$ and differentiate the outer function normally. And then, we will differentiate the inner function accordingly again and write the answer as the product of the two. For example, if there is a function like, $f\left( {g\left( x \right)} \right)$, then we will assume $g\left( x \right)$ as the variable $t$ and differentiate $f\left( t \right)$ and then again differentiate $g\left( x \right)$. A better way to represent this is, $\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f'\left( {g\left( x \right)} \right).g'\left( x \right)$.

Complete step by step answer:
The given function is, $\tan \left( {\sin x} \right)$. Let, $f\left( x \right) = \tan \left( {\sin x} \right)$. Now, differentiating both sides with respect to $x$, we get,
$\dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\tan \left( {\sin x} \right)$
Let, $\sin x = t$. Therefore,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\tan \left( t \right)$
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = {\sec ^2}\left( t \right).\dfrac{{dt}}{{dx}}$
Now, $t = \sin x$. Therefore,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = {\sec ^2}\left( {\sin x} \right).\dfrac{{d\left( {\sin x} \right)}}{{dx}}$
$ \therefore \dfrac{d}{{dx}}f\left( x \right) = {\sec ^2}\left( {\sin x} \right).\cos x$

Therefore, the value of differentiation of $\tan \left( {\sin x} \right)$ is, $\cos x.{\sec ^2}\left( {\sin x} \right)$.

Note: The chain rule is one of the most basic yet most useful rules for differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. The thing we must care for is the correct sequence of choosing the functions and differentiating individually.