This is an exam practice question.
Suppose that $f(x, y)$ is a function of two variables with $f_x(0, 2) = 2$ and $f_y(0, 2) = -1$. Using the chain rule, compute the numerical value of $f_\theta(r\cos\theta, r\sin\theta)$ at $r=2$, $\theta=\pi/2$.
So for this question, I have $x = r\cos\theta$ and $y = r\sin\theta$.
Using the chain rule, I have: \[ f_\theta(r\cos\theta, r\sin\theta) = f_x(x,y)\frac d{d\theta}(r\cos\theta) + f_y(x,y)\frac d{d\theta}(r\sin\theta) \] \[ f_\theta(2\cos\pi/2, 2\sin\pi/2) = (2)(-2\sin\pi/2) - 1(2\cos\pi/2) = -4 \] Do I have the right idea here? Or am I totally off?