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

Last updated date: 13th Jul 2024
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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.