Using implicit function differentiation, find $\left( {\dfrac{{dy}}{{dx}}} \right)$of $\sin x + \cos y = 0$.

Hint:In the given problem, we are required to differentiate $\sin x + \cos y = 0$ with respect to x. Since, $\sin x + \cos y = 0$ is an implicit function, we will have to differentiate the function $\sin x + \cos y = 0$ with the implicit method of differentiation. So, differentiation of $\sin x + \cos y = 0$ with respect to x will be done layer by layer using the chain rule of differentiation as in the given function, we cannot isolate the variables x and y.

Complete step by step answer:
Consider,
$\sin x + \cos y = 0$.
Differentiating both sides of the equation with respect to x, we get,
$\dfrac{d}{{dx}}(\sin x) + \dfrac{d}{{dx}}\left( {\cos y} \right) = \dfrac{d}{{dx}}\left( 0 \right)$
We know that derivative of $\sin \left( x \right)$with respect to x is $\cos \left( x \right)$ and derivative of $\cos \left( y \right)$ with respect to y is $\left( { - \sin \left( y \right)} \right)$.
Hence, we have to apply the chain rule of differentiation in order to differentiate $\cos \left( y \right)$ with respect to x,
 $ = \cos x + ( - \sin y).\dfrac{{dy}}{{dx}} = 0$
Taking $\cos \left( x \right)$ to the right hand side so as to isolate the $\dfrac{{dy}}{{dx}}$ term, we get,
$ = - \sin y\dfrac{{dy}}{{dx}} = - \cos x$
Taking $\left( { - \sin \left( y \right)} \right)$ to the right side of the equation, we get,
$ = \dfrac{{dy}}{{dx}} = \dfrac{{ - \cos x}}{{ - \sin y}}$
Cancelling the negative signs in numerator and denominator, we get,
$ = \dfrac{{dy}}{{dx}} = \dfrac{{\cos x}}{{\sin y}}$
So, the derivative of $\sin x + \cos y = 0$ is $\dfrac{{dy}}{{dx}} = \dfrac{{\cos x}}{{\sin y}}$.
Note: Implicit functions are those functions that involve two variables and the two variables are not separable and cannot be isolated from each other. Hence, we have to follow a certain method to differentiate such functions and also apply the chain rule of differentiation. We must remember the simple derivatives of basic functions to solve such problems.