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.