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Answer 1

Step-by-step explanation:

We have,

$log( \frac{ {x}^{2} + {y}^{2} }{ {x}^{2} - {y}^{2} } ) = {e}^{k}$

$log( {x}^{2} + {y}^{2} ) - log( {x}^{2} - {y}^{2} ) = {e}^{k}$

Differentiating both sides,

$= > \frac{2x + 2y \frac{dy}{dx} }{ {x}^{2} + {y}^{2} } - \frac{2x - 2y \frac{dy}{dx} }{ {x}^{2} - {y}^{2} } = 0$

$= > \frac{2x}{ {x}^{2} + {y}^{2} } + \frac{2y}{ {x}^{2} + {y}^{2} } \frac{dy}{dx} - \frac{2x}{ {x}^{2} - {y}^{2} } + \frac{2y}{ {x}^{2} - {y}^{2} } \frac{dy}{dx} = 0$

$= > 2x( \frac{ {x}^{2} - {y}^{2} - {x}^{2} - {y}^{2} }{ ({x}^{2} + {y}^{2} ) ( {x}^{2} - {y}^{2} ) } ) + 2y\frac{ {x}^{2} - {y}^{2} + {x}^{2} + {y}^{2} }{( {x}^{2} + {y}^{2} )( {x}^{2} } - {y}^{2)} ) \frac{dy}{dx} = 0$

$= > - \frac{4x {y}^{2} }{ {x}^{4} - {y}^{4} } + \frac{4y {x}^{2} }{ {x}^{4} - {y}^{4} } \frac{dy}{dx} = 0$

$= > \frac{4y {x}^{2} }{ {x}^{4} - {y}^{4} } \frac{dy}{dx} = \frac{4x {y}^{2} }{ {x}^{4} - {y}^{4} }$

$= > \frac{dy}{dx} = \frac{y}{x}$