Nabla operation to cos and sin replacement

Question

There are given equations,

$$ x = x' \cos \theta - y'\sin \theta \\ y = x' \sin \theta + y'\cos \theta $$

$$ \frac{\partial f}{\partial x'} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial x'} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x'} $$

I don't understand how it is equal to $$ \frac{\partial f}{\partial x'} = \frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta $$

Can you explain?

Answer 1

$x$ and $y$ are functions that depend on $x'$ and $y'$, i.e., $$x(x',y') = x' \cos\theta-y' \sin\theta,\\ y(x',y') = x' \sin\theta+y' \cos\theta. $$

So \begin{align} \frac{\partial x}{\partial x'} &= \frac{\partial}{\partial x'}(x' \cos\theta-y' \sin\theta) = \frac{\partial}{\partial x'}(x' \cos\theta)\\ &=\cos \theta. \end{align}

Similarly, $\frac{\partial y}{\partial x'} = \sin \theta.$

This is why it is equal to your last equation.