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**Hint:**We are given x = y (log xy). For calculating the derivative of y with respect to x, we will first use the logarithmic relation log (xy) = log x + log y and then we will differentiate with respect to x. on further simplification, we will get the value of $\dfrac{{dy}}{{dx}}$.

**Complete step by step answer:**

We are given that x = y (log xy).

We need to calculate $\dfrac{{dy}}{{dx}}$.

First of all, we will use the logarithmic identity log (xy) = log x + log y in the given equation. On substituting, we get

$ \Rightarrow x = y\left( {\log x + \log y} \right)$

Now, we will differentiate this equation with respect to x and we will use the product rule for differentiating y (log x + log y) which is defined as $\dfrac{{d\left( {mn} \right)}}{{dx}} = n\dfrac{{dm}}{{dx}} + m\dfrac{{dn}}{{dx}}$ . On differentiating w. r. t. x, we get

$ \Rightarrow 1 = \dfrac{{dy}}{{dx}}\left( {\log x + \log y} \right) + y\left( {\dfrac{1}{x} + \dfrac{1}{y}\dfrac{{dy}}{{dx}}} \right)$

$

\Rightarrow 1 = \dfrac{{dy}}{{dx}}\left( {\log x + \log y} \right) + \dfrac{y}{x} + \dfrac{{dy}}{{dx}} \\

\Rightarrow 1 - \dfrac{y}{x} = \dfrac{{dy}}{{dx}}\left( {1 + \log x + \log y} \right) \\$

By using the logarithmic identity log (xy) = log x + log y again

$ \Rightarrow \dfrac{{x - y}}{x} = \dfrac{{dy}}{{dx}}\left( {1 + \log xy} \right) \\

\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x - y}}{{x\left( {1 + \log xy} \right)}} \\

$

Therefore, $\dfrac{{dy}}{{dx}}$ is found to be $\dfrac{{x - y}}{{x\left( {1 + \log xy} \right)}}$.

**Note:**In such problems you can get confused while using various identities and also when you will differentiate a product of two functions. You should be handy of all the identities.

**Additional Information:**In mathematics, many logarithmic identities are there with whom we solve the various problems based on logarithmic functions.

In mathematics, the derivative $\dfrac{{dy}}{{dx}}$ is a function that characterizes the rate of change of the function y with respect to x. The process of finding the derivatives is called differentiation.

Derivatives are fundamentals to the solution of problems on equations in calculus.

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