What is the derivative of ${\tan ^4}\left( {3x} \right)$?

Hint: In the given problem, we are required to differentiate ${\tan ^4}\left( {3x} \right)$ with respect to $x$. The given function is a composite function, so we will have to apply the chain rule of differentiation in the process of differentiation. So, differentiation of ${\tan ^4}\left( {3x} \right)$ with respect to x will be done layer by layer. Also the derivative of $\tan (x)$ with respect to x must be remembered.

Complete step by step answer:
So, Derivative of ${\tan ^4}\left( {3x} \right)$ with respect to $x$ can be calculated as $\dfrac{d}{{dx}}\left( {{{\tan }^4}\left( {3x} \right)} \right)$. Now, $\dfrac{d}{{dx}}\left( {{{\tan }^4}\left( {3x} \right)} \right)$. Now, Let us assume $u = \tan \left( {3x} \right)$. So substituting $\tan \left( {3x} \right)$ as $u$, we get,
$\Rightarrow \dfrac{d}{{dx}}\left( {{u^4}} \right)$
Now, we know that the derivative of ${x^n}$ with respect to $x$ is \[n{x^{n - 1}}\]. So, we get,
$\Rightarrow 4{u^3} \times \dfrac{{du}}{{dx}}$
Now, putting back $u$as $\tan \left( {3x} \right)$, we get,
$\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times \dfrac{{d\left( {\tan \left( {3x} \right)} \right)}}{{dx}}$
Now, let $t = \left( {3x} \right)$. Then, doing the substitution, we get,
$\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times \dfrac{{d\left( {\tan t} \right)}}{{dx}}$
Derivative of tangent function $\tan x$ with respect to x is ${\sec ^2}x$. So, we have,
$\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times {\sec ^2}t \times \dfrac{{dt}}{{dx}}$
Now, putting back t as $\left( {3x} \right)$, we get,
$\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times {\sec ^2}\left( {3x} \right) \times \dfrac{{d\left( {3x} \right)}}{{dx}}$ because \[\dfrac{{dt}}{{dx}} = \dfrac{{d\left( {3x} \right)}}{{dx}}\]
We use the power rule of differentiation $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ again. So, we get,
$\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times {\sec ^2}\left( {3x} \right) \times 3$
Simplifying the product of terms, we get,
$\therefore 12{\tan ^3}\left( {3x} \right){\sec ^2}\left( {3x} \right)$

Hence, the derivative of ${\tan ^4}\left( {3x} \right)$ with respect to $x$ is $12{\tan ^3}\left( {3x} \right){\sec ^2}\left( {3x} \right)$.

Note: This type of composite function will be easily solved by the chain rule of 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. We should always substitute back the value of variables in order to reach the final required answer.