Answer
$f_x=y x^{y-1}$
and
$f_y=x^{y} \ln (x)$
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:
$f_x=\dfrac{\partial }{\partial x} (x^y)=y x^{y-1}$
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:
$f_y=\dfrac{\partial }{\partial y}(x^y) =x^{y} \ln (x)$
