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I came across the following problem:

Given $\displaystyle f(r,\theta)=(r \cos \theta,r \sin \theta)$ for $(r,\theta) \in \mathbb R^2$ with $r \neq 0$. Then how can I find the value of $Df$? ($Df$ denotes the derivative of $f$). Also, how can I check whether $\displaystyle f$ is $1-1$ on $\{(r,\theta) \in \mathbb R^2: r \neq 0\}$ or not?

EDIT: I want to rephrase the first question. I have to check whether the following statement is true/false?

The linear transformation $Df(r,\theta)$ is not zero for any $(r,\theta) \in \mathbb R^2$ with $ r \neq 0$ .

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up vote 1 down vote accepted

Denote $$f_1(r,\; \theta)=r \cos \theta,\\ f_2(r,\; \theta)=r \sin \theta,$$ so $$\displaystyle f(r,\;\theta)=(r \cos \theta,\;r \sin \theta)=(f_1(r,\; \theta),\;f_2(r,\; \theta)).$$ Then $$Df(r,\;\theta)=\begin{pmatrix} \dfrac{\partial{f_1(r,\; \theta)}}{\partial{r}} && \dfrac{\partial{f_1(r,\; \theta)}}{\partial{\theta}} \\ \dfrac{\partial{f_2(r,\; \theta)}}{\partial{r}} && \dfrac{\partial{f_2(r,\; \theta)}}{\partial{\theta}} \end{pmatrix}=\\ =\begin{pmatrix} \dfrac{\partial{(r \cos \theta)}}{\partial{r}} && \dfrac{\partial{(r \cos \theta)}}{\partial{\theta}} \\ \dfrac{\partial{(r \sin \theta)}}{\partial{r}} && \dfrac{\partial{(r \sin \theta)}}{\partial{\theta}} \end{pmatrix}= \begin{pmatrix} \cos{\theta} && -r\sin{\theta} \\ \sin{\theta} && r\cos{\theta} \end{pmatrix} $$ Value of the derivative on vector $\pmatrix{h_1\\h_2}$ equals $$Df(r,\;\theta)\pmatrix{h_1\\h_2}=\begin{pmatrix} \cos{\theta} && -r\sin{\theta} \\ \sin{\theta} && r\cos{\theta} \end{pmatrix}\pmatrix{h_1\\h_2}=\pmatrix{h_1 \cos{\theta}-h_2 r\sin{\theta} \\ h_1\sin{\theta} + h_2r\cos{\theta}}$$

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