What is the derivative of $arctan (\frac{y}{x})$ $arctan(y/x)$ ?

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What is the derivative of $arctan (\frac{y}{x})$ $arctan(y/x)$ ?

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Answer 1 · 2015-08-10T03:04:13

$- \frac{y}{x^{2} + y^{2}} \frac{d y}{d x}$ $- y/(x^2 + y^2) dy/dx$

Explanation:

$\frac{d}{d x} arctan (\frac{y}{x}) = \frac{1}{1 + {(\frac{y}{x})}^{2}} \cdot (\frac{d}{d x} \frac{y}{x})$ $d/(dx) arctan(y/x) = 1/(1 + (y/x)^2) * (d/dx y/x)$
$\frac{d}{d x} arctan (\frac{y}{x}) = \frac{1}{1 + (\frac{y^{2}}{x^{2}})} \cdot (- \frac{y}{x^{2}} \cdot \frac{d y}{d x})$ $d/(dx) arctan(y/x) = 1/(1 + (y^2/x^2)) * (- y/x^2 * dy/dx)$
$\frac{d}{d x} arctan (\frac{y}{x}) = - \frac{x^{2}}{x^{2} + y^{2}} \cdot \frac{y}{x^{2}} \cdot \frac{d y}{d x}$ $d/(dx) arctan(y/x) = - x^2/(x^2 + y^2) * y/x^2 * dy/dx$
$\frac{d}{d x} arctan (\frac{y}{x}) = - \frac{y}{x^{2} + y^{2}} \frac{d y}{d x}$ $d/(dx) arctan(y/x) = - y/(x^2 + y^2) dy/dx$

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What is the derivative of $arctan (\frac{y}{x})$ $arctan(y/x)$ ?

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