How do I solve a partial derivative $f(x,y)$ where $g(r,\theta)=f(r \cos \theta, r \sin \theta)$?

Question

We are given the following information:

1. $f(x,y)$ is differentiable.

2. $g(r, \theta) = f(r \cos \theta, r \sin \theta)$

3. $f_x(1,1)=-2$ and $f_y(1,1)=3$.

We are told to use the chain rule to find $g_r (\sqrt{2},\pi / 4)$.

Attempt 1: (INCORRECT)

My first attempt to solve this involved me thinking about the vector created by the points $(1,1,-2)$ and $(1,1,3)$, which is just a straight line. I figure this implies that $x$ and $y$ are constants.

The next step I took was to integrate $f_x$ and $f_y$. Since $f_x(1,1)=-2 \implies f=-2x+c_y$ and $f_y(1,1)=3 \implies 3y+c_x$. Because the partial derivatives are from the same equation, we can say that $f(x,y)=-2x+3y$. Because $f_r=f_x \cos \theta + f_y \sin \theta$, and the partial derivatives are constants, $g_r (\sqrt{2},\pi / 4) = (-2) \cos \dfrac{\pi}{4} + 3 \sin \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}$.

Answer 1

Second attempt at question, following guidance from the community.

Were given that $\dfrac{\partial f}{\partial x}=-2$ and $\dfrac{\partial f}{\partial y} = 3$

$f_r=f_x \cos \theta + f_y \sin \theta \\ \dfrac{\partial f}{\partial x} \cdot \dfrac{\partial x}{\partial r} + \dfrac{\partial f}{\partial y} \cdot \dfrac{\partial y}{\partial r}=(-2)\left( \cos \dfrac{\pi}{4}\right) + (3) \left( \sin \dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}$

Answer 2

I'll set $\theta=\phi$ for typographical convenience. Now, we have that $g_r=f_r.$ Also, setting $x=r\cos\phi$ and $y=r\sin\phi,$ we have that $$g_r=f_r=f_xx_r+f_yy_r.$$

Also note that when $r=\sqrt 2,\,\phi=π/4,$ we have that $(x,y)=(1,1).$

Can you now complete this?

Answer 3

Your second attempt is basically correct, but there's a term missing from the left side of your second equation. This should be $$ \frac{\partial f}{ \partial x}\cdot\frac{\partial x}{ \partial r}+\frac{\partial f}{ \partial y}\cdot\frac{\partial y}{ \partial r}\ . $$