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What is the derivative of the inverse tan(y/x)?

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May 24, 2015

The derivative would be $\frac{1}{\sqrt{x^{2} + y^{2}}} (\frac{d y}{d x} - \frac{y}{x})$ $1/sqrt(x^2+y^2) (dy/dx -y/x)$

If u is ${tan}^{- 1} (\frac{y}{x})$ $tan^-1(y/x)$ then tan u = $\frac{y}{x}$ $y/x$ . Differentiating w.r.t. x,

${sec}^{2} u \frac{d u}{d x} = \frac{1}{x^{2}} (x \frac{d y}{d x} - y)$ $sec^2u (du)/dx= 1/x^2 (xdy/dx -y)$

$\frac{d u}{d x} = {cos}^{2} u [\frac{1}{x^{2}} (x \frac{d y}{d x} - y)]$ $(du)/dx= cos^2 u [1/x^2(x dy/dx -y)]$

= $\frac{x}{\sqrt{x^{2} + y^{2}}}$ $x/sqrt(x^2+y^2)$ $\frac{1}{x^{2}} (x \frac{d y}{d x} - y)$ $1/x^2 (xdy/dx -y)$

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

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