If y = x operatorname{Tan}^{-1}left(frac{x}{y | vedclass

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Question

(C) Given the equation $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}\right)$.Divide by $x$: $\frac{y}{x} = \operatorname{Tan}^{-1}\left(\frac{x}{y}\

Vedclass · Accepted Answer

$\frac{y}{x}$

Vedclass · Answer

(C) Given the equation $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}ight)$.Divide by $x$: $\frac{y}{x} = \operatorname{Tan}^{-1}\left(\frac{x}{y}ight)$.Let $v = \frac{y}{x}$,so $y = vx$. Then $\frac{x}{y} = \frac{1}{v}$.The equation becomes $v = \operatorname{Tan}^{-1}\left(\frac{1}{v}ight)$,which implies $	an(v) = \frac{1}{v}$,or $v 	an(v) = 1$.However,differentiating $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}ight)$ with respect to $x$ using the product rule and chain rule:$\frac{dy}{dx} = \operatorname{Tan}^{-1}\left(\frac{x}{y}ight) + x \cdot \frac{1}{1 + (x/y)^2} \cdot \frac{d}{dx}\left(\frac{x}{y}ight)$$\frac{dy}{dx} = \operatorname{Tan}^{-1}\left(\frac{x}{y}ight) + x \cdot \frac{y^2}{x^2+y^2} \cdot \left(\frac{y - x \frac{dy}{dx}}{y^2}ight)$$\frac{dy}{dx} = \frac{y}{x} + \frac{x}{x^2+y^2} \cdot (y - x \frac{dy}{dx})$$\frac{dy}{dx} = \frac{y}{x} + \frac{xy}{x^2+y^2} - \frac{x^2}{x^2+y^2} \frac{dy}{dx}$$\frac{dy}{dx} \left(1 + \frac{x^2}{x^2+y^2}ight) = \frac{y}{x} + \frac{xy}{x^2+y^2}$$\frac{dy}{dx} \left(\frac{x^2+y^2+x^2}{x^2+y^2}ight) = \frac{y(x^2+y^2) + x^2y}{x(x^2+y^2)}$$\frac{dy}{dx} \left(\frac{2x^2+y^2}{x^2+y^2}ight) = \frac{yx^2+y^3+x^2y}{x(x^2+y^2)} = \frac{2x^2y+y^3}{x(x^2+y^2)}$$\frac{dy}{dx} = \frac{y(2x^2+y^2)}{x(x^2+y^2)} \cdot \frac{x^2+y^2}{2x^2+y^2} = \frac{y}{x}$.

If $y = x \operatorname{Tan}^{-1}\left(\frac{x}{y}\right)$,then $\frac{dy}{dx} = $

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