Several Variable Calculus & Differential Equations

Question

Suppose that $f(x, y) = g(r, θ)$, where $x = r \cos θ$ and $y = r \sin θ$. Find formulas for $\frac{∂f}{∂x}$ and $\frac{∂f}{∂y}$ which are expressed entirely in terms of $r,θ$ and $\frac{∂g}{∂r}$ and $\frac{∂g}{∂θ}$.

We know that $$(1): \ r^2=x^2+y^2$$ and $$(2):\ θ=\tan^{-1}(\frac xy)$$ So there is a hint on how to do this so i started by finding $\frac{∂g}{∂r}$ and $\frac{∂g}{∂θ}$ using the chain rule. $\frac{∂g}{∂r}=\frac{∂g}{∂x} \frac{∂x}{∂r}+\frac{∂g}{∂y} \frac{∂y}{∂r}=cos^2θ+sin^2θ=1$ and the same way i get $\frac{∂g}{∂θ}=\frac{∂g}{∂x} \frac{∂x}{∂θ}+\frac{∂g}{∂y} \frac{∂y}{∂θ}=0.$

I don't know what i am doing wrong or what i can otherwise. Can someone help?

Answer 1

$$\partial f / \partial x=\partial g / \partial x$$ $$=\partial g / \partial r.\partial r/ \partial x+\partial g / \partial \theta.\partial \theta / \partial x$$ Now take your equation (1) and differentiate both sides with respect to x. Solve for $\partial r/ \partial x$. Differentiate your equation (2) with respect to x. Substitute. Find $\partial f/ \partial y$ similarly.

Answer 2

This is routinely done using the chain rule for Frechet derivatives: $$\frac{\partial f}{\partial (x, y)}(x,y) = \frac{\partial g}{\partial(r, \theta)}(r, \theta)\frac{\partial(r, \theta)}{\partial(x, y)}(x, y) = \frac{\partial g}{\partial(r, \theta)}(r, \theta)\left(\frac{\partial(x, y)}{\partial(r, \theta)}(r, \theta)\right)^{-1}.$$ Written out fully, $$\begin{pmatrix}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y}\end{pmatrix}(x, y) = \begin{pmatrix}\frac{\partial g}{\partial r} & \frac{\partial g}{\partial \theta}\end{pmatrix}(r, \theta) \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\\ \end{pmatrix}(r, \theta)^{-1}.$$