Question on Functions $\mathbb{R}^n \to \mathbb{R} ^m$ I'm having trouble solving this problem:


*

*Let $\bar{x_1} = x_1 \cos(x_2)$ and $ \bar{x_2} = x_1 \sin(x_2)$
Suppose that $f:\mathbb{R^2} \to \mathbb{R^2}$ is a smooth function of $\bar{x_1}$ and $\bar{x_2}.$ Show that: $(\frac{\partial{f}}{\partial\bar{x_1}})^2+(\frac{\partial{f}}{\partial\bar{x_2}})^2 = (\frac{\partial{f}}{\partial{x_1}})^2+ \frac{1}{x_1^2}(\frac{\partial{f}}{\partial{x_2}})^2$.


My Solution:
Using the Chain Rule: $\frac{\partial{f}}{\partial{x_1}} = \frac{\partial{f}}{\partial{\bar{x_1}}} \frac{\partial{\bar{x_1}}}{\partial{x_1}}  + \frac{\partial{f}}{\partial{\bar{x_2}}}\frac{\partial{\bar{x_2}}}{\partial{x_1}}$
 and $\frac{\partial{f}}{\partial{x_2}} = \frac{\partial{f}}{\partial{\bar{x_1}}} \frac{\partial{\bar{x_1}}}{\partial{x_2}}  + \frac{\partial{f}}{\partial{\bar{x_2}}}\frac{\partial{\bar{x_2}}}{\partial{x_2}}$
I can get $\frac{\partial{\bar{x_1}}}{\partial{x_1}} = \cos{x_2}$ and $\frac{\partial{\bar{x_2}}}{\partial{x_1}}=  \sin{x_2}$ and $\frac{\partial{\bar{x_1}}}{\partial{x_2}} = -x_{1}\sin{x_2}$ and $\frac{\partial{\bar{x_2}}}{\partial{x_2}} = x_{1}\cos{x_2}$
But I don't know what else to do; more specifically, I don't know how I am supposed to take the partial derivatives when I don't have the function $f$. How can I solve this problem?
 A: You've basically got the answer here. You have
\begin{equation}\tag{*}
\partial_{x_{1}}f = \cos x_{2}\partial_{\overline{x}_{1}}f + \sin x_{2}\partial_{\overline{x}_{2}}f\\
\partial_{x_{2}}f = -x_{1}\sin x_{2}\partial_{\overline{x}_{1}}f + x_{1}\cos x_{2} \partial_{\overline{x}_{2}}f. 
\end{equation}
Compute $(\partial_{x_{1}}f)^{2} + \frac{1}{x_{1}^{2}}(\partial_{x_{2}}f)^{2}$ and see what you get.
Edit: $(\partial_{x_{1}}f)^{2} = \cos^{2}x_{2}(\partial_{\overline{x}_{1}}f)^{2} + \sin^{2}x_{2}(\partial_{\overline{x}_{2}}f)^{2} + 2\cos x_{2}\sin x_{2}\partial_{\overline{x}_{1}}f\partial_{\overline{x}_{2}}f$.
You can compute $\frac{1}{x_{1}^{2}}(\partial_{x_{2}}f)^{2}$ similarly and then add them. You don't need to compute explicit partials of some specific function. In this case, you only need to know how the partials relate, which the chain rule tells you.
Edit 2: From the application of the chain rule that you included in your question we get $(*)$. This implies that
$$(\partial_{x_{1}}f)^{2} = (\cos x_{2}\partial_{\overline{x}_{1}}f + \sin x_{2}\partial_{\overline{x}_{2}}f)^{2} = \cos^{2}x_{2}(\partial_{\overline{x}_{1}}f)^{2} + \sin^{2}x_{2}(\partial_{\overline{x}_{2}}f)^{2} + 2\cos x_{2}\sin x_{2}\partial_{\overline{x}_{1}}f\partial_{\overline{x}_{2}}f.$$
Similarly,
\begin{align*}
\frac{1}{x_{1}^{2}}(\partial_{x_{2}}f)^{2} &= \frac{1}{x_{1}^{2}}(x_{1}^{2}\sin^{2}x_{2}(\partial_{\overline{x}_{1}}f)^{2} + x_{1}^{2}\cos^{2}x_{2}(\partial_{\overline{x}_{2}}f)^{2} - 2 x_{1}^{2}\cos x_{2}\sin x_{2}\partial_{\overline{x}_{1}}f\partial_{\overline{x}_{2}}f)\\
&= \sin^{2}x_{2}(\partial_{\overline{x}_{1}}f)^{2} + \cos^{2}x_{2}(\partial_{\overline{x}_{2}}f)^{2} - 2\cos x_{2}\sin x_{2}\partial_{\overline{x}_{1}}f\partial_{\overline{x}_{2}}f.
\end{align*}
Now if you add the above expressions together and simplify just a bit you should get the desired equality.
