How do you differentiate $f\left( x \right)=\left( \dfrac{1}{{{x}^{3}}} \right)\sin x$ using the product rule?

Hint: From the question given we have the function $f\left( x \right)=\left( \dfrac{1}{{{x}^{3}}} \right)\sin x$ and now we have to find the differentiation of the function by using product rule. Now we have to differentiate both sides with respect to the x then we have to use UV rule in the function while differentiating the right hand side part. UV rule means when we differentiate any function in the form UV then we will have to differentiate like \[D\left( UV \right)=D\left( U \right)\times V+D\left( V \right)\times U\]. we will use the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}}$

Complete step-by-step solution:
From the question, given function is
$\Rightarrow f\left( x \right)=\left( \dfrac{1}{{{x}^{3}}} \right)\sin x$
Now, we have to differentiate both sides with respect to the x.
By differentiating both sides with respect to x we will get,
$\Rightarrow f\left( x \right)=\left( \dfrac{1}{{{x}^{3}}} \right)\sin x$
$\Rightarrow {{f}^{|}}\left( x \right)=\dfrac{d}{dx}\left( \left( \dfrac{1}{{{x}^{3}}} \right)\sin x \right)$
Here the right-hand side part $\left( \dfrac{1}{{{x}^{3}}} \right)\sin x$ is in the form of UV.
We know that if any function is in the form of UV then we will use UV rule for differentiating the function.
UV rule means,
\[\Rightarrow D\left( UV \right)=D\left( U \right)\times V+D\left( V \right)\times U\]
Here U is $\left( \dfrac{1}{{{x}^{3}}} \right)$ and V is \[\sin x\].
We will write $\left( \dfrac{1}{{{x}^{3}}} \right)$ as ${{x}^{-3}}$
Then by substituting in the above rule we will get,
$\Rightarrow {{f}^{|}}\left( x \right)=\dfrac{d}{dx}\left( {{x}^{-3}} \right)\times \sin x+\dfrac{d}{dx}\left( \sin x \right)\times {{x}^{-3}}$
We know that differentiation of ${{x}^{-3}}$ with respect to x is $\left( -3\times {{x}^{-4}} \right)$.
By using formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}}$
$\Rightarrow {{f}^{|}}\left( x \right)=\left( -3\times {{x}^{-4}} \right)\times \sin x+\dfrac{d}{dx}\left( \sin x \right)\times {{x}^{-3}}$
We know that differentiation of \[\sin x\] with respect to x is \[\cos x\].
Then we will get,
$\begin{align}
  & \Rightarrow {{f}^{|}}\left( x \right)=\left( -3\times {{x}^{-4}} \right)\times \sin x+\left( \cos x \right)\times {{x}^{-3}} \\
 & \\
\end{align}$
$\Rightarrow {{f}^{|}}\left( x \right)=\dfrac{x\cos x-3\sin x}{{{x}^{4}}}$
Therefore, we want the differential${{f}^{|}}\left( x \right)$ by using product rule is ,
$\Rightarrow {{f}^{|}}\left( x \right)=\dfrac{x\cos x-3\sin x}{{{x}^{4}}}$

Note: Students should know the basic differentiation formulas of differentiation. Students should be very careful while using the formula $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\times {{x}^{n-1}}$ here in the above question the n values is $-3$ not $3$. If we take $3$ we will get the wrong answer and the whole solution will be wrong.