# Find $\dfrac{{dy}}{{dx}}$where${x^y} = {y^x};{\text{ x > 0,y > 0}}$.

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HInt: here you can take log both sides and then use rules of differentiation to make it easy.

We have to find the derivative of ${x^y} = {y^x};{\text{ x > 0,y > 0}}$

So let’s take log both sides we get

$\log ({x^y}) = \log ({y^x})$

Using the property of logarithm that$\log ({a^b}) = b\log a$, we can write above as

$y \times \log x = x \times \log y$

Now let’s differentiate both the sides with respect to $x$ using the product rule of derivative and chain rule we have

$\dfrac{{dy}}{{dx}} \times \log x + y \times \dfrac{1}{x} = 1 \times \log y + x \times \dfrac{1}{y}\dfrac{{dy}}{{dx}}$

Let’s take $\dfrac{{dy}}{{dx}}$terms to left hand side

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

So our $\dfrac{{dy}}{{dx}}$is

$\dfrac{{dy}}{{dx}} = \dfrac{{\log y - \dfrac{y}{x}}}{{\log x - \dfrac{x}{y}}}$

$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{x\log y - y}}{x}}}{{\dfrac{{y\log x - x}}{y}}}$

Let’s simplify it further

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

Note-The key concept to solve such a problem statement is to take logarithm both sides and apply the property of log as mentioned above before differentiating , this always will take you to the right answer.

We have to find the derivative of ${x^y} = {y^x};{\text{ x > 0,y > 0}}$

So let’s take log both sides we get

$\log ({x^y}) = \log ({y^x})$

Using the property of logarithm that$\log ({a^b}) = b\log a$, we can write above as

$y \times \log x = x \times \log y$

Now let’s differentiate both the sides with respect to $x$ using the product rule of derivative and chain rule we have

$\dfrac{{dy}}{{dx}} \times \log x + y \times \dfrac{1}{x} = 1 \times \log y + x \times \dfrac{1}{y}\dfrac{{dy}}{{dx}}$

Let’s take $\dfrac{{dy}}{{dx}}$terms to left hand side

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

So our $\dfrac{{dy}}{{dx}}$is

$\dfrac{{dy}}{{dx}} = \dfrac{{\log y - \dfrac{y}{x}}}{{\log x - \dfrac{x}{y}}}$

$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{x\log y - y}}{x}}}{{\dfrac{{y\log x - x}}{y}}}$

Let’s simplify it further

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

Note-The key concept to solve such a problem statement is to take logarithm both sides and apply the property of log as mentioned above before differentiating , this always will take you to the right answer.

Last updated date: 18th Sep 2023

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