Как решать такие примеры [tex]\frac{dy}{dx}[/tex] к примеру [tex]\frac{y}{x}=arctg(xy)[/tex]. Created by мороженкаAikas. matematika-ru

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July 2022 1 11 Report

Как решать такие примеры [tex]\frac{dy}{dx}[/tex]
к примеру [tex]\frac{y}{x}=arctg(xy)[/tex]

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hello93

$\frac{y}{x} = \arctan(xy) \\ \frac{d}{dx} ( \frac{y}{x} ) = \frac{d}{dx} ( \arctan(xy)) \\ \frac{y'x - y}{ {x}^{2} } = \frac{y + xy'}{1 + {x}^{2} {y}^{2} } \\ \frac{y'}{x} - \frac{y}{ {x}^{2} } = \frac{y}{ 1 + {x}^{2} {y}^{2} } + \frac{xy'}{1 + {x}^{2} {y}^{2} } \\ \frac{y'}{x} - \frac{xy'}{1 + {x}^{2} {y}^{2} } = \frac{y}{ {x}^{2} } + \frac{y}{1 + {x}^{2} {y}^{2} } \\ y'( \frac{1}{x} - \frac{x}{1 + {x}^{2} {y}^{2} } ) = \frac{y}{ {x}^{2} } + \frac{y}{1 + {x}^{2} {y}^{2} } \\ y' = \frac{ \frac{y}{ {x}^{2} } + \frac{y}{1 + {x}^{2} {y}^{2} } }{ \frac{1}{x} - \frac{x}{1 + {x}^{2} {y}^{2} } } = \frac{ \frac{y(1 + {x}^{2} {y}^{2}) + y {x}^{2} }{ {x}^{2}(1 + {x}^{2} {y}^{2} )} }{ \frac{1 + {x}^{2} {y}^{2} - {x}^{2} }{x(1 + {x}^{2} {y}^{2} )} } = \frac{y(1 + {x}^{2} {y}^{2} + {x}^{2}) }{x(1 + {x}^{2} {y}^{2} - {x}^{2} )}$

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