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 $$ u = x\; cos \theta - y\; sin \theta $$ $$ v = x\; sin \theta + y\; cos\theta $$ 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.
$$ df = \left(\frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} \right) dx + \left(\frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} \right) dy $$