Question

find dy/dx of implicit derivative y=x^2+3y^2+y^3-x​

Answer 1

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Answer 2

Step-by-step explanation:

$\sf y = {x}^{2} + 3 {y}^{2} + {y}^{3} - x$

$\sf differentiating \:w. r. t. \: x$

$\sf \frac {dy}{dx} = 2x + 6y \frac {dy}{dx} + 3{y}^{2} \frac {dy}{dx} - 1$

$\sf \frac {dy}{dx} ( 1 - 6y - 3 {y}^{2} ) = 2x - 1$

$\sf \frac {dy}{dx} = \frac {2x - 1 }{ 1 - 6y - 3 {y}^{2}}$

$\boxed {\bold {\sf \frac {dy}{dx} = \frac {2x - 1 }{ 1 - 6y - 3 {y}^{2}}}}$

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