Answer
$-6 sin (3x-y^2) cos (3x-y^2) ; \\ 4y sin (3x-y^2) cos (3x-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 keeping $y$ as a constant, and vice versa:
$f_x=2 \cos (3x-y^2) \times [-\sin (3x-y^2)] \times \dfrac{\partial (3x-y^2)}{\partial x}=-6 sin (3x-y^2) cos (3x-y^2) $
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=2 \cos (3x-y^2) \times [-\sin (3x-y^2)] \times \dfrac{\partial (3x-y^2)}{\partial y}=4y sin (3x-y^2) cos (3x-y^2) $
