Question

Find the differential of the function. v = 2y cos(xy)

Answer 1

Given a function f(x,y), the differential of f is given by
$df = f_x dx + f_y dy$
where $f_x$ is the derivative of f(x,y) with respect to x, while $f_y$ is the derivative of f(x,y) with respect to y.

The function of our problem is$f(x,y)=2y \cos(xy)$
The derivative in x is
$f_x = 2y(-y \sin (xy))=-2y^2 \sin (xy)$
The derivative in y is
$f_y = 2 \cos (xy) + 2y (-x \sin (xy))=2 \cos (xy) - 2xy \sin (xy)$

So, the differential of f(x,y) is
$df= -2y^2 \sin (xy) dx + (2 \cos (xy) - 2xy \sin (xy))dy$