How do you implicitly differentiate $ \ln (xy) = x + y $

Hint: Here in this question the function is involving the logarithmic function. “ln” represents the logarithmic function. To differentiate the above function first we apply the properties of logarithmic and then we apply the differentiation to the function.

Complete step-by-step answer:
The logarithmic functions are the inverse of exponential function. The given function is of logarithmic function. We use the product rule of logarithmic function, and we apply the differentiation.
Now consider the given function $ \ln (xy) = x + y $ , on applying the product rule of logarithmic function it is defined as $ \ln (xy) = \ln x + \ln y $ . On applying the product rule of logarithmic function, the given function is written as
 $ \Rightarrow \ln x + \ln y = x + y $ ------ (1)
The differentiation is defined as a rate of change of one quantity with respect to other quantity.
Now differentiate the equation (1) with respect to x we get,
 $ \Rightarrow \dfrac{d}{{dx}}(\ln x) + \dfrac{d}{{dx}}(\ln y) = \dfrac{d}{{dx}}(x) + \dfrac{d}{{dx}}(y) $
The differentiation of $ \ln x $ is $ \dfrac{1}{x} $
So, we get,
 $ \Rightarrow \dfrac{1}{x}\dfrac{d}{{dx}}(x) + \dfrac{1}{y}\dfrac{d}{{dx}}(y) = 1 + \dfrac{{dy}}{{dx}} $
On simplification we can write it has
 $ \Rightarrow \dfrac{1}{x} + \dfrac{1}{y}\dfrac{{dy}}{{dx}} = 1 + \dfrac{{dy}}{{dx}} $
Let we move $ \dfrac{{dy}}{{dx}} $ to LHS and $ \dfrac{1}{x} $ to RHS so we get
 $ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} - \dfrac{{dy}}{{dx}} = 1 - \dfrac{1}{x} $
We take $ \dfrac{{dy}}{{dx}} $ as common in the LHS.
 $ \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{1}{y} - 1} \right) = 1 - \dfrac{1}{x} $
Tale LCM on both sides we have
 $ \Rightarrow \dfrac{{dy}}{{dx}}\left( {\dfrac{{1 - y}}{y}} \right) = \dfrac{{x - 1}}{x} $
On the further simplification the above equation is written as
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y(x - 1)}}{{x(1 - y)}} $
The above equation can be written as
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{xy - y}}{{x - xy}} $
Hence, we obtained the derivative.
Therefore, the implicit differentiation of $ \ln (xy) = x + y $ is $ \dfrac{{dy}}{{dx}} = \dfrac{{xy - y}}{{x - xy}} $
So, the correct answer is “ $ \dfrac{{dy}}{{dx}} = \dfrac{{xy - y}}{{x - xy}} $ ”.

Note: The logarithmic function is a reciprocal or inverse of an exponential function. We have logarithmic properties for addition, subtraction, product and division. Here in this question, we use the property of product rule for logarithmic function. The differentiation is defined as the rate of change of quantity with respect to other quantities. After applying the property of logarithm then we apply the differentiation to the function.