Find dy dx if x^3 + x^2y + xy^2 + y^3 = 81. - Vidyadip

Question

Find $\frac{{dy}}{{dx}}$ if $x^3 + x^2y + xy^2 + y^3 = 81$.

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Answer

we have,$x^3 + x^2y + xy^2 + y^3 = 81$
Differentiating both sides w.r.t to x,we get,
$3{x^2} + {x^2}.\frac{{dy}}{{dx}} + y.2x + x.2y\frac{{dy}}{{dx}} + {y^2}.1 + 3{y^2}\frac{{dy}}{{dx}} = 0$
$\left( {{x^2} + 2xy + 3{y^2}} \right)\frac{{dy}}{{dx}} = - 3{x^2} - 2xy - {y^2}$
$\frac{{dy}}{{dx}} = \frac{{ - \left( {3{x^2} + 2xy + {y^2}} \right)}}{{{x^2} + 2xy + 3{y^2}}}$

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