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$2x^3 + x^2y-xy^3 = 2$

$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$

$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y6^2 \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$

$\frac{\mathrm{d}y}{\mathrm{d}x} (x^2 +3y^2)(6x^2+2x-y^3) = 0$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6x^2+2x-y^3}{x^2+3y^2} $

Did I tackle this question correctly?

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    @Sri, my last comment was for you, too.2011-10-25

1 Answers 1

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$2x^3 + x^2y-xy^3 = 2$

$\frac{\mathrm{d}}{\mathrm{d}x} [2x^3+x^2y -xy^3 ] = \frac{\mathrm{d}}{\mathrm{d}x}(2)$

$6x^2 + \left(2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x}\right) - \left( 1 y^3 + 3y^2x \frac{\mathrm{d}y}{\mathrm{d}x}\right ) = 0$

$\frac{\mathrm{d}y}{\mathrm{d}x} (x^2 - 3y^2x) + (6x^2+2xy-y^3) = 0$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6x^2+2xy-y^3}{3y^2x -x^2} $

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    I totally missed that but thanks! At least i was on the right track :D2011-10-25