Given a function f(x,y), the differential of f is given by
[tex]df = f_x dx + f_y dy[/tex]
where [tex]f_x[/tex] is the derivative of f(x,y) with respect to x, while [tex]f_y[/tex] is the derivative of f(x,y) with respect to y.
The function of our problem is[tex]f(x,y)=2y \cos(xy)[/tex]
The derivative in x is
[tex]f_x = 2y(-y \sin (xy))=-2y^2 \sin (xy)[/tex]
The derivative in y is
[tex]f_y = 2 \cos (xy) + 2y (-x \sin (xy))=2 \cos (xy) - 2xy \sin (xy)[/tex]
So, the differential of f(x,y) is
[tex]df= -2y^2 \sin (xy) dx + (2 \cos (xy) - 2xy \sin (xy))dy[/tex]
