If z = tan^{-1}left(frac{x}{y}right),then z_x | vedclass

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(C) Given $z = 	an^{-1}\left(\frac{x}{y}ight)$.First,find the partial derivative of $z$ with respect to $x$ $(z_x)$:$z_x = \frac{\partial z}{\parti

Vedclass · Accepted Answer

$-y:x$

Vedclass · Answer

(C) Given $z = 	an^{-1}\left(\frac{x}{y}ight)$.First,find the partial derivative of $z$ with respect to $x$ $(z_x)$:$z_x = \frac{\partial z}{\partial x} = \frac{1}{1 + (x/y)^2} \cdot \frac{1}{y} = \frac{1}{(y^2 + x^2)/y^2} \cdot \frac{1}{y} = \frac{y^2}{x^2 + y^2} \cdot \frac{1}{y} = \frac{y}{x^2 + y^2}$.Next,find the partial derivative of $z$ with respect to $y$ $(z_y)$:$z_y = \frac{\partial z}{\partial y} = \frac{1}{1 + (x/y)^2} \cdot \left(-\frac{x}{y^2}ight) = \frac{y^2}{x^2 + y^2} \cdot \left(-\frac{x}{y^2}ight) = -\frac{x}{x^2 + y^2}$.Now,calculate the ratio $z_x : z_y$:$z_x : z_y = \frac{y}{x^2 + y^2} : -\frac{x}{x^2 + y^2} = y : -x = -y : x$.Therefore,the correct option is $C$.

If $z = \tan^{-1}\left(\frac{x}{y}\right)$,then $z_x : z_y = $

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