# Showing that $\frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy = d\left(\tan^{-1}\left(\frac{x}{y}\right)\right)$

I'm trying to show that

$$\frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy = d\left(\tan^{-1}\left(\frac{x}{y}\right)\right)$$

but am having trouble figuring out exactly how to approach the problem. I've tried a few tricks with switching to polar coordinates or trying to make use of right triangles, but have had no luck. Could anyone point me in the right direction here?

• You don't need polar or right triangles for this. – user_of_math Jul 18 '14 at 13:26
• Do you know the derivative of $\operatorname{tan}^{-1}$? – Giulio Bresciani Jul 18 '14 at 13:27
• This is not exactly true, because the LHS is defined in $\mathbb{R}^2-\{(0,0)\}$, while the RHS is defined in $\{(x,y)\in\mathbb{R}^2:y\neq0\}$. – enzotib Jul 18 '14 at 14:38

Hint: $$\frac{d}{dt}\tan^{-1}(t) = \frac{1}{1+t^2}$$ $$d f(x,y) = \frac{\partial f(x,y)}{\partial x} dx + \frac{\partial f(x,y)}{\partial y} dy$$
\begin{align} \frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy &=\frac{1}{(\frac{x}{y})^2+1}\frac{1}{y}dx+ \frac{1}{(\frac{x}{y})^2+1}(-\frac{x}{y^2}) dy\\ &=\frac{1}{(\frac{x}{y})^2+1}(\frac{1}{y}dx+(-\frac{x}{y^2})dy)\\ &=\frac{1}{(\frac{x}{y})^2+1}d(\frac{x}{y})\\ &= d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) \end{align}