Question

Find $\frac{dy}{dx}$, if $x^3+y^3+4x^{3}y=0$

Answer 1

$x^3+y^3+4x^{3}y=0$

differentiating both sides,

$\Rightarrow 3x^2+3y^2\frac{dy}{dx}+4x^3\frac{dy}{dx}+4y.3x^2=0$

$\Rightarrow 3x^2+(3y^2+4x^3)\frac{dy}{dx}+12x^{2}y=0$

$\Rightarrow \frac{dy}{dx}=\frac{-12x^{2}y-3x^2}{(3y^2+4x^3)}=\frac{-(12x^{2}y+3x^2)}{3y^2+4x^3}$

Hence, the answer is $\frac{-(12x^{2}y+3x^2)}{3y^2+4x^3}.$