Answer
$\dfrac{1}{x \ln y} ; \\ \dfrac{-\ln x}{y(\ln y)^2}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by treating $y$ as a constant, and vice versa:
$f_x=\dfrac{1}{\ln y} [ \dfrac{\partial (\ln x)}{\partial x}]=\dfrac{1}{x \ln y}$
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $y$, by treating $x$ as a constant, and vice versa:
$f_y=[(-\ln y)^2] \ln (x) \dfrac{\partial (\ln y)}{\partial y}=\dfrac{-\ln x}{y(\ln y)^2}$
