Question

Find dy/dx by differentiating implicitly x^2y+3xy^3-x=3

Answer 1

$\dfrac{\mathrm d}{\mathrm dx}\left[x^2y+3xy^3-x\right]=\dfrac{\mathrm d}{\mathrm dx}[3]$
$2xy+x^2\dfrac{\mathrm dy}{\mathrm dx}+3y^3+9xy^2\dfrac{\mathrm dy}{\mathrm dx}-1=0$
$(x^2+9xy^2)\dfrac{\mathrm dy}{\mathrm dx}=1-2xy-3y^3$
$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{1-2xy-3y^3}{x^2+9xy^2}$