How do you implicitly differentiate \[\sin x + \cos y = \sin x\cos y\] ?

Hint: Differentiation calculates the rate of change of a given quantity. Implicit function definition conveys when we are not able to isolate the dependent variable in a differential equation becomes an implicit function. Both the dependent variable and independent variables are present in this type of function.

Complete step-by-step answer:
Given, \[\sin x + \cos y = \sin x\cos y\] .
Doing implicit differentiation, applying differentiation on both side,
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x + \cos y} \right) = \dfrac{d}{{dx}}\left( {\sin x\cos y} \right)\]
Applying sum or difference rule on the left hand side of the differential equation and Product rule on the right hand side of the differential equation we have,
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x} \right) + \dfrac{d}{{dx}}\left( {\cos y} \right) = \dfrac{d}{{dx}}\left( {\sin x\cos y} \right)\]
Product rule we have \[f'(x) = u'(x).v(x) + u(x).v'(x)\] . Where \[u(x) = \sin x\] and \[v(x) = \cos y\] .
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x} \right) + \dfrac{d}{{dx}}\left( {\cos y} \right) = \dfrac{d}{{dx}}(\sin x)\cos y + \sin x\dfrac{d}{{dx}}(\cos y)\]
We know that, \[\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x\] and \[\dfrac{d}{{dx}}(\cos y) = - \sin y\dfrac{{dy}}{{dx}}\] . Substituting this we have,
 \[ \Rightarrow \cos x + \left( { - \sin y\dfrac{{dy}}{{dx}}} \right) = \cos x\cos y + \sin x\left( { - \sin y\dfrac{{dy}}{{dx}}} \right)\]
 \[ \Rightarrow \cos x - \sin y\dfrac{{dy}}{{dx}} = \cos x\cos y - \sin y\sin x\dfrac{{dy}}{{dx}}\] .
Regrouping \[\dfrac{{dy}}{{dx}}\] terms we get,
 \[ \Rightarrow \sin y\sin x\dfrac{{dy}}{{dx}} - \sin y\dfrac{{dy}}{{dx}} = \cos x\cos y - \cos x\]
Taking \[\dfrac{{dy}}{{dx}}\] common on the left hand side of the equation,
 \[ \Rightarrow \dfrac{{dy}}{{dx}}\left( {\sin y\sin x - \sin y} \right) = \cos x\cos y - \cos x\]
Now dividing \[\left( {\sin y\sin x - \sin y} \right)\] on both sides of the problem we have,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos x\cos y - \cos x}}{{\sin y\sin x - \sin y}}\]
Taking \[\cos x\] common in the numerator and taking \[\sin y\] common in the denominator term we have,
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos x(\cos y - 1)}}{{\sin y(\sin x - 1)}}\] . This is the required answer.
So, the correct answer is “ \[ \dfrac{{dy}}{{dx}} = \dfrac{{\cos x(\cos y - 1)}}{{\sin y(\sin x - 1)}}\] ”.

Note: We have different types of differentiation rules.
Sum or difference rule: when the function is the sum or difference of two functions, the derivative is the sum or difference of derivative of each function. i.e.,
 \[f(x) = u(x) \pm v(x)\] then \[f'(x) = u'(x) \pm v'(x)\] .
Product rule: when f(x) is the product of two function that is \[f(x) = u(x).v(x)\] then \[f'(x) = u'(x).v(x) + u(x).v'(x)\] .
Quotient rule: , if the two functions \[f(x)\] and \[g(x)\] are differentiable then the quotient is differentiable and \[{\left( {\dfrac{{f(x)}}{{g(x)}}} \right)^1} = \dfrac{{f'(x)g(x) - f(x)g'(x)}}{{{{\left( {g(x)} \right)}^2}}}\] .
We use this rule depending on the given problem. Careful in the differentiation a function f(y) with respect to x. we will have \[\dfrac{d}{{dx}}(f(y)) = f'(y)\dfrac{{dy}}{{dx}}\] .