Question

find the derivatives of:x^y=y^x​

Answer 1

$\bold {xy \: . \: yx}$

$\bold{Taking \: log \: on \: both \: sides - }$

$\bold{y \: logx \: = \: x \: log y.}$

$\bold{Differentiating \: both \: the \: sides \: by \: uv \: rule}$

$\bold{⇒ \:y. \dfrac{1}{x} + log_{x} \: . \dfrac{dy}{dx} = x. \dfrac{1}{y} \: \dfrac{dy}{dx} + log_{y}} \\ \\ \bold{⇒ \frac{y}{x} - log_{y} = \frac{x}{y} . \frac{dy}{dx} - log_{x} \: \frac{dy}{dx} } \\ \\ \bold{⇒ \: \frac{y - x \: log_{y} }{x} \: = \: \frac{dy}{dx} \: ( \dfrac{x - y \: log_{x} }{y}) } \\ \\ \bold{⇒ \: \frac{dy}{dx} \: = \: \dfrac{ \dfrac{y - x \: log_{y} }{x} }{ \dfrac{x - y \: log_{x} }{y} } } \\ \\ \bold{⇒ \frac{dy}{dx} \: = \: \frac{x}{y} \: \: (\frac{y - x \: log_{y} }{x - y \: log_{x} } )}$