How do you find the derivative of $y=\dfrac{1}{x}-3\sin x$?

Hint: In this question we have been given a function which is equal to the difference of the two functions $\dfrac{1}{x}$ and $3\sin x$. So we need to differentiate these two functions separately. The derivative of the function ${{x}^{n}}$ is equal to $n{{x}^{n-1}}$. Also, the derivative of the function $\sin x$ is equal to $\cos x$. With the help of these, we can find out the derivative of both of the functions and the final derivative of the given function will be equal to the difference between the derivatives of these functions.

Complete step-by-step solution:
The given function in the question is
$y=\dfrac{1}{x}-3\sin x........(i)$
Let us suppose that \[f\left( x \right)=\dfrac{1}{x}\] and \[g\left( x \right)=3\sin x\]
So the equation (i) can be written as
$\Rightarrow y=f\left( x \right)-g\left( x \right)$
Differentiating both the sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right) \right)}{dx}-\dfrac{d\left( g\left( x \right) \right)}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=f'\left( x \right)-g'\left( x \right)........(ii) \\
\end{align}\]
So we need to differentiate the functions $f\left( x \right)$ and $g\left( x \right)$ separately.
According to our assumption, the function $f\left( x \right)$ is
$\begin{align}
  & \Rightarrow f\left( x \right)=\dfrac{1}{x} \\
 & \Rightarrow f\left( x \right)={{x}^{-1}} \\
\end{align}$
Differentiating both the sides, we get
$\Rightarrow f'\left( x \right)=\dfrac{d\left( {{x}^{-1}} \right)}{dx}$
We know that the derivative of the function ${{x}^{n}}$ is equal to $n{{x}^{n-1}}$. Putting $n=-1$, we get the derivative of the function ${{x}^{-1}}$ equal to $-{{x}^{-1-1}}$. Putting it in the above equation, we get
\[\Rightarrow f'\left( x \right)=-{{x}^{-2}}.........(iii)\]
Now, the function $g\left( x \right)$ is
$\Rightarrow g\left( x \right)=3\sin x$
Differentiating both sides, we get
\[\begin{align}
  & \Rightarrow g'\left( x \right)=\dfrac{d\left( 3\sin x \right)}{dx} \\
 & \Rightarrow g'\left( x \right)=3\dfrac{d\left( \sin x \right)}{dx} \\
\end{align}\]
We know that the derivative of the function $\sin x$ is equal to $\cos x$. Putting it in the above equation, we get
\[\Rightarrow g'\left( x \right)=3\cos x...........(iv)\]
Substituting the equations (iii) and (iv) in (ii), we get
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=-{{x}^{-2}}-3\cos x \\
 & \Rightarrow \dfrac{dy}{dx}=-\left( {{x}^{-2}}+3\cos x \right) \\
\end{align}$
Hence, the derivative of $y=\dfrac{1}{x}-3\sin x$ is equal to $-\left( {{x}^{-2}}+3\cos x \right)$.

Note: The derivative of the function $\sin x$ is equal to $\cos x$ and is not equal to the negative of $\cos x$. This confusion might occur since the integration of $\sin x$ is equal to the negative of $\cos x$. We must not forget this difference between the derivative and the integration formulae.