If x^2 tan ^{-1} frac{y}{x}-y^2 tan ^{-1} fra | vedclass

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(C) Given equation is $x^2 	an ^{-1}\left(\frac{y}{x}ight)-y^2 	an ^{-1}\left(\frac{x}{y}ight)=k$. Differentiating both sides with respect to $x

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(C) Given equation is $x^2 	an ^{-1}\left(\frac{y}{x}ight)-y^2 	an ^{-1}\left(\frac{x}{y}ight)=k$. Differentiating both sides with respect to $x$: $x^2 \cdot \frac{1}{1+(y/x)^2} \cdot \frac{x(dy/dx)-y}{x^2} + 2x 	an ^{-1}\left(\frac{y}{x}ight) - y^2 \cdot \frac{1}{1+(x/y)^2} \cdot \frac{y-x(dy/dx)}{y^2} - 2y 	an ^{-1}\left(\frac{x}{y}ight) \frac{dy}{dx} = 0$. Simplifying the terms: $\frac{x^2}{x^2+y^2} \cdot \frac{x(dy/dx)-y}{1} + 2x 	an ^{-1}\left(\frac{y}{x}ight) - \frac{y^2}{x^2+y^2} \cdot \frac{y-x(dy/dx)}{1} - 2y 	an ^{-1}\left(\frac{x}{y}ight) \frac{dy}{dx} = 0$. At the point $(1,1)$,we have $x=1, y=1$ and $dy/dx = y_1$: $\frac{1}{2}(y_1-1) + 2(1) 	an ^{-1}(1) - \frac{1}{2}(1-y_1) - 2(1) 	an ^{-1}(1) y_1 = 0$. Note that $	an ^{-1}(1) = \pi/4$. $\frac{1}{2}y_1 - \frac{1}{2} + 2(\pi/4) - \frac{1}{2} + \frac{1}{2}y_1 - 2(\pi/4)y_1 = 0$. $y_1 - 1 + \pi/2 - \pi/2 y_1 = 0$. $y_1(1 - \pi/2) = 1 - \pi/2$. Thus,$y_1 = 1$.

If $x^2 \tan ^{-1} \frac{y}{x}-y^2 \tan ^{-1} \frac{x}{y}=k$,then $\left(\frac{d y}{d x}\right)_{(1,1)}=$

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