let $x(\theta)=r (\cos(\theta))$ and $y(\theta)=r(\sin(\theta))$ in $f(x(\theta),y(\theta))$, write $\frac{\partial f}{\partial x} \times \frac{\partial f}{\partial y}$ in terms of $\theta$ and $r$.

Any help would be appreciated.

-
Do you mean the cross product of those two vector functions? –  Henry Shearman Dec 11 '11 at 1:18
@HenryShearman - No –  Victor Dec 11 '11 at 1:21
Ah yes, I see what you mean now. $f\,$ is a scalar function. –  Henry Shearman Dec 11 '11 at 1:27
But what is $f$? You have no hope of computing its partial derivatives without knowing the function itself. And is $r$ somehow constant while $\theta$ is a variable, or what? –  Henning Makholm Dec 11 '11 at 1:31

## closed as not a real question by Victor, Austin Mohr, t.b., Asaf Karagila, QuixoticDec 11 '11 at 14:32

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I assume you mean $x(r,\theta) = r \cos(\theta)$ and $y(r,\theta) = r \sin(\theta)$.

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial x}$$

$$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial f}{\partial \theta} \frac{\partial \theta}{\partial y}$$

We have $r^2 = x^2 + y^2$ and $\tan(\theta) = \frac{y}{x}$.

Hence, $$r \frac{\partial r}{\partial x} = x \implies \frac{\partial r}{\partial x} = \cos(\theta)$$ $$r \frac{\partial r}{\partial y} = y \implies \frac{\partial r}{\partial y} = \sin(\theta)$$ $$\sec^2(\theta) \frac{\partial \theta}{\partial x} = -\frac{y}{x^2} \implies \frac{\partial \theta}{\partial x} = -\frac{r \sin(\theta)}{r^2 \cos^2(\theta) \sec^2(\theta)} = - \frac{\sin(\theta)}{r}$$ $$\sec^2(\theta) \frac{\partial \theta}{\partial y} = \frac1{x} \implies \frac{\partial \theta}{\partial y} = \frac1{r \cos(\theta) \sec^2(\theta)} = \frac{\cos(\theta)}{r}$$ Hence, you get $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \cos(\theta) - \frac{\partial f}{\partial \theta} \frac{\sin(\theta)}{r}$$

$$\frac{\partial f}{\partial y} = \frac{\partial f}{\partial r} \sin(\theta) + \frac{\partial f}{\partial \theta} \frac{\cos(\theta)}{r}$$

I would like to point out that $$\frac{\partial \theta}{\partial x} \neq \frac1{\frac{\partial x}{\partial \theta}}$$ $$\frac{\partial \theta}{\partial y} \neq \frac1{\frac{\partial y}{\partial \theta}}$$ $$\frac{\partial r}{\partial x} \neq \frac1{\frac{\partial x}{\partial r}}$$ $$\frac{\partial r}{\partial y} \neq \frac1{\frac{\partial y}{\partial r}}$$

For more details on this, look up here.

-