Question

$find \: \frac{dy}{dx} if \\ {x}^{2} + {xy}^{3} + {x}^{2} y + {y}^{4 = 4}$
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Answer 1

${\bold{\huge{\underline{\pink{\underline{ANSWER}:-}}}}}$

${x}^{2} + x {y}^{3} + {x}^{2}y + {y}^{4} = 4 \\ = > \frac{d( {x}^{2}) }{dx} + \frac{d(x {y}^{3}) }{dx} + \frac{d( {x}^{2}y) }{dx} + \frac{d({y}^{4}) }{dx} = 4 \\ = > please \: see \: the \: attachment \: picture \: for \: details \: explanations. \\ = > \frac{dy}{dx} = - ( \frac{2x + {y}^{3} + 2xy }{ {x}^{2} + 4 {y}^{3} + 3x {y}^{2} } )$

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Answer 2

ANSWER​:−​

$\begin{lgathered}{x}^{2} + x {y}^{3} + {x}^{2}y + {y}^{4} = 4 \\ = > \frac{d( {x}^{2}) }{dx} + \frac{d(x {y}^{3}) }{dx} + \frac{d( {x}^{2}y) }{dx} + \frac{d({y}^{4}) }{dx} = 4 \\ = > please \: see \: the \: attachment \: picture \: for \: details \: explanations. \\ = > \frac{dy}{dx} = - ( \frac{2x + {y}^{3} + 2xy }{ {x}^{2} + 4 {y}^{3} + 3x {y}^{2} } )\end{lgathered}x2+xy3+x2y+y4=4=>dxd(x2)​+dxd(xy3)​+dxd(x2y)​+dxd(y4)​=4=>pleaseseetheattachmentpicturefordetailsexplanations.=>dxdy​=−(x2+4y3+3xy22x+y3+2xy​)​$

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