Question

If $ x = y\ln \left( {xy} \right) $ , then $ \dfrac{{dy}}{{dx}} $ equals to:
(A) $ \dfrac{{y\left( {x - y} \right)}}{{x\left( {x + y} \right)}} $
(B) $ \dfrac{{x\left( {x + y} \right)}}{{y\left( {x - y} \right)}} $
(C) $ \dfrac{{y\left( {x + y} \right)}}{{x\left( {x - y} \right)}} $
(D) $ \dfrac{{x\left( {x - y} \right)}}{{y\left( {x + y} \right)}} $

Answer 1

Hint: In the given problem, we are required to differentiate the given function in x and y: $ x = y\ln \left( {xy} \right) $ with respect to x. Since, $ x = y\ln \left( {xy} \right) $ is an implicit function, so we will have to differentiate the function $ x = y\ln \left( {xy} \right) $ with the implicit method of differentiation. So, differentiation of $ x = y\ln \left( {xy} \right) $ 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:
So, we have, $ x = y\ln \left( {xy} \right) $ .
Differentiating both sides of the equation with respect to x, we get,
 $ \Rightarrow \dfrac{d}{{dx}}\left( x \right) = \dfrac{d}{{dx}}\left( {y\ln \left( {xy} \right)} \right) $
We know that the derivative of $ x $ with respect to x is $ 1 $ and derivative of $ \ln \left( x \right) $ with respect to x is $ \left( {\dfrac{1}{x}} \right) $ .
We also know the product rule of differentiation $ \dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) $ .
So, applying product rule in the right side of the equation, we get,
 $ \Rightarrow 1 = y\dfrac{d}{{dx}}\left( {\ln \left( {xy} \right)} \right) + \ln \left( {xy} \right)\dfrac{{dy}}{{dx}} $
Hence, we have to apply the chain rule of differentiation in order to differentiate $ \ln \left( {xy} \right) $ with respect to x,
  $ \Rightarrow 1 = \dfrac{y}{{xy}}\left( {\dfrac{{d\left( {xy} \right)}}{{dx}}} \right) + \ln \left( {xy} \right)\dfrac{{dy}}{{dx}} $
Again using the product rule of differentiation in order to simplify the expression,
 $ \Rightarrow 1 = \dfrac{y}{{xy}}\left( {x\dfrac{{dy}}{{dx}} + y\left( 1 \right)} \right) + \ln \left( {xy} \right)\dfrac{{dy}}{{dx}} $
Cancelling the common factors in numerator and denominator, we get,
 $ \Rightarrow 1 = \dfrac{1}{x}\left( {x\dfrac{{dy}}{{dx}} + y\left( 1 \right)} \right) + \ln \left( {xy} \right)\dfrac{{dy}}{{dx}} $
Now, opening the brackets, we get,
\[ \Rightarrow 1 = \dfrac{{dy}}{{dx}} + \dfrac{y}{x} + \ln \left( {xy} \right)\dfrac{{dy}}{{dx}}\]
Keeping all the terms consisting $ \dfrac{{dy}}{{dx}} $ on the right side and shifting the rest of the terms on the left side of the equation, we get
\[ \Rightarrow 1 - \dfrac{y}{x} = \dfrac{{dy}}{{dx}}\left( {1 + \ln \left( {xy} \right)} \right)\]
\[ \Rightarrow \dfrac{{1 - \dfrac{y}{x}}}{{\left( {1 + \ln \left( {xy} \right)} \right)}} = \dfrac{{dy}}{{dx}}\]
Taking LCM, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x - y}}{{x\left( {1 + \ln \left( {xy} \right)} \right)}}\]
But the options given to us in the question don't have this expression. So, we would have to simplify the expression to match the options.
The given expression is $ x = y\ln \left( {xy} \right) $ .
Evaluating the value of $ \ln \left( {xy} \right) $ , we get,
 $ \Rightarrow \ln \left( {xy} \right) = \dfrac{x}{y} $
So, putting the value of $ \ln \left( {xy} \right) $ in \[\dfrac{{dy}}{{dx}} = \dfrac{{x - y}}{{x\left( {1 + \ln \left( {xy} \right)} \right)}}\], we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x - y}}{{x\left( {1 + \dfrac{x}{y}} \right)}}\]
Taking LCM, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x - y}}{{x\left( {\dfrac{{y + x}}{y}} \right)}}\]
Simplifying the expression further, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y\left( {x - y} \right)}}{{x\left( {y + x} \right)}}\]
So, the derivative of $ x = y\ln \left( {xy} \right) $ is \[\dfrac{{y\left( {x - y} \right)}}{{x\left( {y + x} \right)}}\].
Hence, the option (A) is correct.
So, the correct answer is “Option A”.

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.