Ytukyg

How come
$$frac{partial f}{partial x} = frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x}$$
when
$$
u = x; cos theta - y; sin theta
$$
$$
v = x; sin theta + y; costheta
$$
It seems like that formula is twice as great on the right side. (Are the definitions of $u$ and $v$ even relevant?)

ANSWER: This is the definition of the chain rule for partial derivatives when $f$ is a function of $x$ and $y.$ It is irrelevant to the definitions of $u(x,y)$ and $v(x,y).$

This is not a proof, but a demonstration: Had $f=f(u,v)$ only where u and v were independent variables, you'd get a total change in f from changes in both $u$ and $v$

$$
df = frac{partial f}{partial u} du + frac{partial f}{partial v} dv
$$
I suspect you are comfortable with that. As it turns out, $u$ and $v$ are themselves functions of independent variables x and y (assuming $theta$ is only a parameter here). So

$$
du = frac{partial u}{partial x} dx + frac{partial u}{partial y} dy \
dv = frac{partial v}{partial x} dx + frac{partial v}{partial y} dy
$$

Substitute for these values of $du,dv$
$$
df = frac{partial f}{partial u} (frac{partial u}{partial x} dx + frac{partial u}{partial y} dy ) + frac{partial f}{partial v} (frac{partial v}{partial x} dx + frac{partial v}{partial y} dy )
$$

Group terms in dx, dy together and you have your result.

$$
df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
$$