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If x = y (log xy), then find $\dfrac{{dy}}{{dx}}$.

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Last updated date: 13th Jun 2024
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Answer
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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.