Question

$\rm Find \: \frac{dy}{dx} : -$
$\rm {x}^{y} + {y}^{x} = 1$
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Answer 1

${x}^{y} + {y}^{x} = 1 \\ take \: differentiation \: \\ to \: both \: side \\ \frac{dy}{dx} ( {x}^{y} + {y}^{x} ) = \frac{dy}{dx} \\ \frac{d}{dx} y {x}^{y} + \frac{d}{dx} yy {}^{x} \\ \frac{d}{dx} y = y \frac{d}{dx} {x}^{y} + {x}^{y} \frac{d}{dx} y + \frac{d}{dx} {y}^{x + 1} \\ \frac{d}{dx}y = y \times( y {x}^{y - 1} ) + {x}^{y} \frac{d}{dx}y +$

Answer 2

Answer:

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Step-by-step explanation:

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