How do you differentiate ${\sin ^3}4x$?

Hint: To differentiate ${\sin ^3}4x$ first substitute $\sin x$ to some variable $t$ and then use the chain rule of differentiation i.e. $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$. In the next step, again use the chain rule of differentiation to differentiate $\sin 4x$ by substituting $4x$ to some other variable $u$. Use simple formulas $\dfrac{d}{{dx}}\sin x = \cos x,{\text{ }}\dfrac{d}{{dx}}{x^3} = 3{x^2}{\text{ and }}\dfrac{d}{{dx}}4x = 4$ in simplification to get the final answer.

Complete step by step answer:
According to the question, we have to show the differentiation process of ${\sin ^3}4x$.
Let this function be denoted by $y$. So we have:
$ y = {\sin ^3}4x$
To differentiate this function, we will use the chain rule of differentiation. First, let’s substitute $\sin 4x = t$. So our function will become:
$ y = {t^3}$
Now, according to the chain rule of differentiation, we have:
$ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}}$
Applying this rule for our function, we’ll get:
$ \dfrac{{dy}}{{dx}} = \dfrac{d}{{dt}}\left( {{t^3}} \right) \times \dfrac{{dt}}{{dx}}$
We know that the differentiation of ${t^3}$ with respect to $t$ is $3{t^2}$. Using this we will get:
 $ \dfrac{{dy}}{{dx}} = 3{t^2} \times \dfrac{{dt}}{{dx}}$
Putting back the value of $t$, we will get:
$ \dfrac{{dy}}{{dx}} = 3{\left( {\sin 4x} \right)^2} \times \dfrac{d}{{dx}}\left( {\sin 4x} \right){\text{ }}.....{\text{(1)}}$
Now, to differentiate $\sin 4x$ we’ll again use substitution and substitute $4x = u$ and if we apply chain rule again, we have:
$ \dfrac{d}{{dx}}\left( {\sin 4x} \right) = \dfrac{d}{{du}}\left( {\sin u} \right) \times \dfrac{{du}}{{dx}}$
Putting this in our differentiation i.e. equation (1), we’ll get:
$ \dfrac{{dy}}{{dx}} = 3{\sin ^2}4x \times \dfrac{d}{{du}}\left( {\sin u} \right) \times \dfrac{{du}}{{dx}}$
We know that the differentiation of $\sin u$ with respect to $u$ is $\cos u$. Using this we will get:
$ \dfrac{{dy}}{{dx}} = 3{\sin ^2}4x \times \cos u \times \dfrac{{du}}{{dx}}$
Putting back the value of $u$, we have:
$ \dfrac{{dy}}{{dx}} = 3{\sin ^2}4x \times \cos 4x \times \dfrac{d}{{dx}}4x$
Further, we know that the differentiation of $4x$ with respect to $x$ is $4$. Putting this we will get:
$
 \dfrac{{dy}}{{dx}} = 3{\sin ^2}4x \times \cos 4x \times 4 \\
\Rightarrow \dfrac{{dy}}{{dx}} = 12{\sin ^2}4x\cos 4x \\
 $

Thus the differentiation of ${\sin ^3}4x$ with respect to $x$ is $12{\sin ^2}4x\cos 4x$.

Note: Whenever we have to differentiate a composite function, we always use the chain rule of differentiation after substitution. This makes a complex looking function simple from where we can differentiate step by step. For example, consider the given composite function:
$ y = f\left( {g\left( x \right)} \right)$
To differentiate this function, we’ll substitute $g\left( x \right) = t$, we will have:
$ y = f\left( t \right)$
Now we can apply chain rule of differentiation as shown below:
$ \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}f\left( t \right) \times \dfrac{{dt}}{{dx}}$
Now this differentiation is simple and we can do it step by step. After doing this, we can put back the value of $t$ to get the answer.