Solve the given differentiation $\dfrac{d}{{dx}}\left( {\dfrac{x}{{\sin x}}} \right)$
a)${\text{csc }}x\left( {1 + x\cot x} \right)$
b)${\text{csc }}x\left( {x\cot x - 1} \right)$
c)${\text{csc }}x\left( {1 - x\cot x} \right)$
d)${\text{csc }}x\left( {1 + \cos x} \right)$

Hint: We can use quotient rule to solve this differentiation which is given as-
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{g^2}\left( x \right)}}$
Where $f\left( x \right)$ and $g\left( x \right)$ are functions of x and $f'\left( x \right)$ is first order derivative of function$f\left( x \right)$ while $g'\left( x \right)$ is first order derivative of $g\left( x \right)$.Then use the formula$\dfrac{1}{{\sin x}} = \csc x$ and $\dfrac{{\cos x}}{{\sin x}} = \cot x$ and simplify the equation.

Complete step-by-step answer:
We have to differentiate $\dfrac{d}{{dx}}\left( {\dfrac{x}{{\sin x}}} \right)$
So we will use the formula-
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{g^2}\left( x \right)}}$
Assume here $f\left( x \right) = x$ and $g\left( x \right) = \sin x$ then on using the formula we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \dfrac{{\sin x\dfrac{{dx}}{{dx}} - x\dfrac{{d\left( {\sin x} \right)}}{{dx}}}}{{{{\sin }^2}x}}$
Where $f\left( x \right)$ and $g\left( x \right)$ are functions of x and $f'\left( x \right)$ is first order derivative of function$f\left( x \right)$ while $g'\left( x \right)$ is first order derivative of $g\left( x \right)$
Now we know that $\dfrac{{d\sin x}}{{dx}} = \cos x$ and differentiation of x is $1$
So on using these formulas we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \dfrac{{\sin x - x\cos x}}{{{{\sin }^2}x}}$
On separating the terms we get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \dfrac{{\sin x}}{{{{\sin }^2}x}} - \dfrac{{x\cos x}}{{{{\sin }^2}x}}\]
On solving further we get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \dfrac{1}{{\sin x}} - \dfrac{{x\cos x}}{{{{\sin }^2}x}}\]
And we know that $\dfrac{1}{{\sin x}} = \csc x$
So we can write,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \csc x - \dfrac{{x\cos x}}{{{{\sin }^2}x}}\]
Now we can also write the second term as-
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \csc x - \dfrac{x}{{\sin x}}.\dfrac{{\cos x}}{{\sin x}}\]
And we know that $\dfrac{1}{{\sin x}} = \csc x$ and $\dfrac{{\cos x}}{{\sin x}} = \cot x$
So on putting these values in the equation we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \csc x - x\csc x.\cot x$
Now here we see that $\csc x$ is a common term so we take it common from the equation and we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \csc x\left( {1 - x\cot x} \right)$
Hence the correct answer is C.

Note: We can also solve the given question this way-
Given, $\dfrac{d}{{dx}}\left( {\dfrac{x}{{\sin x}}} \right)$
We know that$\dfrac{1}{{\sin x}} = \csc x$ so we can write –
$ \Rightarrow $ $\dfrac{d}{{dx}}\left( {x\csc x} \right)$
Now on using chain rule we get,
$ \Rightarrow \csc x\dfrac{d}{{dx}}x + x\dfrac{d}{{dx}}\left( {\csc x} \right)$
And we know that,$\dfrac{{d(\csc x)}}{{dx}} = - \csc x\cot x$ and we know the differentiation of x is one.
So on using this formula we get,
$ \Rightarrow \csc x + x\left( { - \csc x\cot x} \right)$
On simplifying we get,
$ \Rightarrow \csc x - x\csc x\cot x$
Now here we see that $\csc x$ is a common term so we take it common from the equation and we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{x}{{\sin x}}} \right] = \csc x\left( {1 - x\cot x} \right)$
Hence we get the same answer by using this method.