How do you find $\dfrac{{dy}}{{dx}}$ if $x + \tan \left( {xy} \right) = 0$ ?

Hint: In this question, we are given an equation and we have been asked to find its derivative. You will have to use chain rule while answering this question. First, simply find the derivative. While differentiating the second term of the given equation, using chain rule. After that, use some very basic trigonometric formulae to simplify the answer.

Formula used: 1) $\dfrac{{d\left( x \right)}}{{dx}} = 1$
2) $\dfrac{{d\left( {\tan x} \right)}}{{dx}} = {\sec ^2}x$

Complete step-by-step solution:
We are given an equation $x + \tan \left( {xy} \right) = 0$.
Let us differentiate both the sides with respect to x.
$ \Rightarrow x + \tan \left( {xy} \right) = 0$
Differentiating both the sides with respect to x,
$ \Rightarrow \dfrac{d}{{dx}}\left( {x + \tan \left( {xy} \right)} \right) = \dfrac{{d\left( 0 \right)}}{{dx}}$
$ \Rightarrow 1 + {\sec ^2}\left( {xy} \right) \times \dfrac{{d\left( {xy} \right)}}{{dx}} = 0$ …. (Here, we used chain rule to find the derivative of$\tan \left( {xy} \right)$)
Differentiating using chain rule and product rule,
$ \Rightarrow 1 + {\sec ^2}\left( {xy} \right) \times \left[ {x\dfrac{{dy}}{{dx}} + y} \right] = 0$
Opening the brackets,
$ \Rightarrow 1 + x{\sec ^2}\left( {xy} \right)\dfrac{{dy}}{{dx}} + y{\sec ^2}xy = 0$
Shifting all the terms to the RHS, other than the term containing$\dfrac{{dy}}{{dx}}$,
$ \Rightarrow x{\sec ^2}\left( {xy} \right)\dfrac{{dy}}{{dx}} = - \left( {y{{\sec }^2}xy + 1} \right)$
Now, we will shift the co-efficient of $\dfrac{{dy}}{{dx}}$,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \left( {y{{\sec }^2}xy + 1} \right)}}{{x{{\sec }^2}xy}}$
Expanding the RHS,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \left( {\dfrac{{y{{\sec }^2}xy}}{{x{{\sec }^2}xy}} + \dfrac{1}{{x{{\sec }^2}xy}}} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \left( {\dfrac{y}{x} + \dfrac{{{{\cos }^2}xy}}{x}} \right)$ …. $\left( {\dfrac{1}{{\sec x}} = \cos x} \right)$

Hence, $\dfrac{{dy}}{{dx}}$ of $x + \tan \left( {xy} \right) = 0$ is $ - \left( {\dfrac{y}{x} + \dfrac{{{{\cos }^2}xy}}{x}} \right)$.

Note: Chain rule: What is chain rule? This rule is used to find the derivative of composite functions. In a composite function, there is an outer function and an inner function. Correct identification of these functions is very important. Let us understand chain rule with the help of an example.
For example: $\tan \left( {xy} \right)$ is a composite function, where $\tan x$ is an outer function and $xy$ is an inner function. How do we find its derivative? At first, we differentiate the outer function and assume the inner function to be x (if we are differentiating with respect to x). In the next step, we differentiate the inner function.
Hence, correct identification of outer and inner function correctly is very important.