How do you differentiate $f\left( x \right) = \dfrac{{\cos x}}{{\sin x}}$ ?

Hint: In this question, we have been given a trigonometric equation and we have been asked to differentiate the equation. This question can be done in two ways.
Method 1: Use quotient rule to find the differentiation. Once you have put the values in the quotient rule, you will see that a trigonometric formula will be used here. Put that formula in the equation and you will have your answer.
Method 2: Use trigonometry to simplify the given ratio. You will get a trigonometric ratio. Simply, write its differentiation and you will get the answer. But, if you are bad at remembering formulas, refer to the method 1 above.

Formula used: 1) Quotient rule: $\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right).f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$
2) $\dfrac{{d\left( {\cot x} \right)}}{{dx}} = \cos e{c^2}x$

Complete step-by-step solution:
We are given a trigonometric equation and we have to find its differentiation. I will use method 1 here.
In this method, we will simply use quotient rules. The quotient rule says that –
$\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$
$ \Rightarrow f\left( x \right) = \dfrac{{\cos x}}{{\sin x}}$ …. (given)
Differentiating both the sides with respect to x by applying the quotient rule,
$ \Rightarrow f'\left( x \right) = \dfrac{{\cos x\left( {\sin x} \right)' - \sin x\left( {\cos x} \right)'}}{{{{\sin }^2}x}}$
Simplifying the equation,
$ \Rightarrow f'\left( x \right) = \dfrac{{\cos x\cos x + \sin x\sin x}}{{{{\sin }^2}x}}$
$ \Rightarrow f'\left( x \right) = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\sin }^2}x}}$
Now, we know that ${\cos ^2}x + {\sin ^2}x = 1$. Using this in the equation,
$ \Rightarrow f'\left( x \right) = \dfrac{1}{{{{\sin }^2}x}}$
We also know that $\dfrac{1}{{\sin x}} = \cos ecx$. Using this to simplify the equation,
$ \Rightarrow f'\left( x \right) = \cos e{c^2}x$

Hence, the differentiation of $\dfrac{{\cos x}}{{\sin x}}$ is $\cos e{c^2}x$.

Note: Now, I will show you how we can solve the given equation using method 2.
$ \Rightarrow f\left( x \right) = \dfrac{{\cos x}}{{\sin x}}$ …. (given)
We know that $\dfrac{{\cos x}}{{\sin x}} = \cot x$. Using this, we get,
$ \Rightarrow f\left( x \right) = \cot x$
Differentiating both the sides with respect to x,
$f'\left( x \right) = \cos e{c^2}x$
Hence, this is our final answer and it is similar to the answer we got using the previous method.