# Question #3c093

Jan 1, 2018

$\frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \cdot \left(\frac{\frac{\mathrm{dy}}{\mathrm{dx}} \cdot x - 1 \cdot y}{x} ^ 2\right)$

#### Explanation:

Use implicit differentiation.
The derivative of $\arctan \left(f \left(x\right)\right)$ is $\frac{1}{1 + {\left(f \left(x\right)\right)}^{2}} \cdot f ' \left(x\right)$ (chain rule)

This means $\frac{d}{\mathrm{dx}} \left(\arctan \left(\frac{y}{x}\right)\right) = \frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{y}{x}\right)$

Using quotient rule:
$= \frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \cdot \left(\frac{\frac{\mathrm{dy}}{\mathrm{dx}} \cdot x - 1 \cdot y}{x} ^ 2\right)$

Jan 1, 2018

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

#### Explanation:

Given: ${\tan}^{-} 1 \left(\frac{y}{x}\right)$

Using the chain rule, where $u = \frac{y}{x}$

$\frac{d \left({\tan}^{-} 1 \left(\frac{y}{x}\right)\right)}{\mathrm{dx}} = \frac{d \left({\tan}^{-} 1 \left(u\right)\right)}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

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

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

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

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

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