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

Hint: We need to find the derivative of the function ${{\sin }^{3}}\left( 4x \right)$ . We start to solve the problem by considering u=4x, v = $\sin 4x$ , and y = ${{\sin }^{3}}\left( 4x \right)$ . Then, we find the derivative of the given function using the formula $\dfrac{dy}{dx}=\dfrac{du}{dx}\times \dfrac{dv}{du}\times \dfrac{dy}{dv}$ .

Complete step by step solution:
We are given a function and need to find the derivative of it. We solve this question using the chain rule in differentiation.
The chain rule is used to find the derivatives of the composite functions.
Let us consider,
$\Rightarrow u=4x$
Differentiating the above equation on both sides with respect to $x$ , we get,
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( 4x \right)$
From the formulae of differentiation, we know that
$\Rightarrow \dfrac{d}{dx}\left( ax \right)=a$
Following the same, we get,
$\therefore \dfrac{du}{dx}=4$
Now, let us consider,
$\Rightarrow v=\sin \left( 4x \right)$
From the above, we know that the value of the variable $u=4x$
Substituting the same, we get,
$\Rightarrow v=\sin u$
Differentiating the above equation on both sides with respect to $u$ , we get,
$\Rightarrow \dfrac{dv}{du}=\dfrac{d}{du}\left( \sin u \right)$
From the formulae of differentiation, we know that the derivative of the sine function is the positive cosine function.
Writing the same in the form of the equation, we get,
$\Rightarrow \dfrac{d}{du}\left( \sin u \right)=\cos u$
Substituting the same in the above equation, we get,
$\Rightarrow \dfrac{dv}{du}=\cos u$
Further, assume the variable y such that
$\Rightarrow y={{\sin }^{3}}\left( 4x \right)$
From the above, we know that the value of the variable $v=\sin 4x$
Substituting the same, we get,
$\Rightarrow y={{v}^{3}}$
Differentiating the above equation on both sides with respect to $v$ , we get,
$\Rightarrow \dfrac{dy}{dv}=\dfrac{d}{dv}\left( {{v}^{3}} \right)$
From the formulae of differentiation, we know that
$\Rightarrow \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$
Following the same, we get,
$\therefore \dfrac{dy}{dv}=3{{v}^{2}}$
The derivative of the given function can be found out using the chain rule as follows,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{du}{dx}\times \dfrac{dv}{du}\times \dfrac{dy}{dv}$
Substituting the values in the above equation, we get,
$\Rightarrow \dfrac{d}{dx}\left( {{\sin }^{3}}\left( 4x \right) \right)=4\times \cos u\times 3{{v}^{2}}$
We know that $u=4x$ and $v=\sin 4x$ . substituting the same, we get,
$\Rightarrow \dfrac{d}{dx}\left( {{\sin }^{3}}\left( 4x \right) \right)=4\times \cos 4x\times 3{{\left( \sin 4x \right)}^{2}}$
Simplifying the above equation, we get,
$\therefore \dfrac{d}{dx}\left( {{\sin }^{3}}\left( 4x \right) \right)=12\cos 4x{{\sin }^{2}}4x$

Note: A composite function is a function that is written inside another function. We must always remember that the derivative of a composite function can be found out using the chain rule of differentiation.