I need to find
$$\frac{\partial Z}{\partial U} \text{ and } \frac{\partial Z}{\partial V}$$
for a $z=f(x,y) = \cos(xy) + y\cos(x)$.
After a bit of an internet search, I think I have found the correct way to obtain a partial derivative for a trigonometric function but I am not so sure at the same time.
Would someone be able to confirm if
$$\frac{\partial Z}{\partial U} = -y\sin(xy) - y\sin(x)$$
And as
$$\ x=u^2+v$$ and $$y=u-v^2$$ would $$\frac{\partial Z}{\partial U} =-y\sin[(u^2-u)(u-v^2)] - y\sin(u^2+v)$$
Which would equal $$\ -y\sin(u^3-uv+v^2+v^2u^2) - y\sin(u^2+v)$$
Any help is appreiciated and warmly recieved! I think once I understand how to find $$\frac{\partial Z}{\partial U}$$ I will be able to find $$ \frac{\partial Z}{\partial V}$$
Hint: For change of variables $(x,y) \to (u,v)$, the chain rule is $$ \begin{align} \frac{\partial f}{\partial u} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial u}\\ \frac{\partial f}{\partial v} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial v} \end{align} $$
Details in chapter 2 here