How do you differentiate $xy = \cot (xy)$?

Hint: Here in this question, we have to find the derivative of a function with respect to x. The function is a trigonometric function. To find the derivative of a function we use the product rule of differentiation. Hence, we obtain the required result for the question.

Complete step by step solution:
The differentiation or derivative of a function is known as rate of change of a quantity. Here we have to differentiate the given function, the trigonometric function.
Now consider the given function
$xy = \cot (xy)$----- (1)
We differentiate the given function with respect to x.
$\dfrac{d}{{dx}}\left( {xy} \right) = \dfrac{d}{{dx}}\left( {\cot (xy)} \right)$
In the LHS of the above equation the term is a product of two terms. So we apply the product of differentiation. The product of differentiation is defined as $\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$, where u and v are functions of x.
So on differentiation we have
$ \Rightarrow x\dfrac{{dy}}{{dx}} + y\dfrac{{dx}}{{dx}} = - \cos e{c^2}(xy)\dfrac{d}{{dx}}(xy)$
Now we will apply the product of differentiation to the term of RHS so we have
$ \Rightarrow x\dfrac{{dy}}{{dx}} + y\dfrac{{dx}}{{dx}} = - \cos e{c^2}(xy)\left[ {x\dfrac{{dy}}{{dx}} + y\dfrac{{dx}}{{dx}}} \right]$
On simplification we have
$ \Rightarrow x\dfrac{{dy}}{{dx}} + y = - \cos e{c^2}(xy)\left[ {x\dfrac{{dy}}{{dx}} + y} \right]$
Let we group the $\dfrac{{dy}}{{dx}}$ on one side, the equation is written as
$ \Rightarrow x\dfrac{{dy}}{{dx}} + x\cos e{c^2}(xy)\dfrac{{dy}}{{dx}} = - y\cos e{c^2}(xy) - y$
Now let we take $\dfrac{{dy}}{{dx}}$ common in LHS and y in RHS, so we have
$ \Rightarrow \dfrac{{dy}}{{dx}}(x + x\cos e{c^2}(xy)) = - y(\cos e{c^2}(xy) + 1)$
In the LHS of the above equation we can take x as common, so we have
$ \Rightarrow x\dfrac{{dy}}{{dx}}(1 + \cos e{c^2}(xy)) = - y(\cos e{c^2}(xy) + 1)$
Now we cancel the same or similar terms which are in both LHS and RHS. The equation is written as
$ \Rightarrow x\dfrac{{dy}}{{dx}} = - y$
Now move the x to the RHS we have
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - y}}{x}$
Hence the differentiation of $xy = \cot (xy)$ is $\dfrac{{ - y}}{x}$

Note: The student must know about the differentiation formulas for the trigonometry ratios and these differentiation formulas are standard. If the function is a product of two terms and the both terms are the function of x then we use the product rule of differentiation to the function.