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 $f_\theta(r\cos\theta,r\sin\theta) = 2$ at $r=2$, $\theta=\frac{\pi}{2}$.
Any hints on how to do this question would be appreciated. Thanks in advance.
Hint: Note that, $f = f(x, y)$ and $x=r\cos(\theta)$, $ y=r\sin(\theta) $ $$ f_{\theta} = f_{x}x_{\theta}+ f_{y}y_{\theta} $$
where $x_{\theta} = \frac{\partial x}{\partial \theta}$ and same for $y_{\theta}$.
The problem is not quite properly written, but you should be able to understand what is meant by the problem.
I can try to make it more rigorous:
Let $f_1$ and $f_2$ be the partial derivatives of $f:\mathbb R^2 \to \mathbb R$ with respect to the first variable and the second variable respectively.
$x(r, \theta) = r\cos\theta$
$y(r, \theta) = r\sin\theta$
$f_1(0, 2) = 2$
$f_2(0, 2) = -1$
$g(r, \theta) = f(x(r, \theta), y(r, \theta))$
Find $g_2(2, \pi/2)$.
To solve this problem, use the chain rule: $$ g_2(r, \theta) = f_1(x(r,\theta),y(r,\theta))x_2(r,\theta) + f_2(x(r,\theta),y(r,\theta))y_2(r,\theta). $$ At $r = 2$ and $\theta = \pi/2$, you should be able to evaluate all terms on the right-hand side.
