# How do you take the derivative of Tan^-1 (y/x)?

Jun 7, 2015

I'm assuming you are thinking of this as being a function of two independent variables $x$ and $y$: $z = {\tan}^{- 1} \left(\frac{y}{x}\right)$. The answers are $\setminus \frac{\setminus \partial z}{\setminus \partial x} = - \setminus \frac{y}{{x}^{2} + {y}^{2}}$ and $\setminus \frac{\setminus \partial z}{\setminus \partial y} = \setminus \frac{x}{{x}^{2} + {y}^{2}}$.

Both of these facts can be derived with the Chain Rule, the Power Rule, and the fact that $\frac{y}{x} = y {x}^{- 1}$ as follows:

$\setminus \frac{\setminus \partial z}{\setminus \partial x} = \setminus \frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \setminus \cdot \setminus \frac{\setminus \partial}{\setminus \partial x} \left(y {x}^{- 1}\right) = \setminus \frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \setminus \cdot \left(- y {x}^{- 2}\right)$

$= - \setminus \frac{y}{{x}^{2} + {y}^{2}}$

and

$\setminus \frac{\setminus \partial z}{\setminus \partial y} = \setminus \frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \setminus \cdot \setminus \frac{\setminus \partial}{\setminus \partial y} \left(y {x}^{- 1}\right) = \setminus \frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \setminus \cdot \left({x}^{- 1}\right)$

$= \setminus \frac{\frac{1}{x}}{1 + {y}^{2} / \left({x}^{2}\right)} = \setminus \frac{x}{{x}^{2} + {y}^{2}}$