## Calculus: Early Transcendentals (2nd Edition)

$\frac{{dy}}{{dx}} = \frac{{1 - 3y\,{{\left( {xy + 1} \right)}^2}}}{{3x\,{{\left( {xy + 1} \right)}^2} + 2y}}$
$\begin{gathered} \,{\left( {xy + 1} \right)^3} = x - {y^2} + 8 \hfill \\ \hfill \\ implicit\,\,differentiation \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {\,{{\left( {xy + 1} \right)}^3}} \right] = \frac{d}{{dx}}\,\,\left[ {x - {y^2} + 8} \right] \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \end{gathered}$ $3\,{\left( {xy + 1} \right)^2}\,\left( {xy' + y} \right) = 1 - 2yy'$ $multiply$ $3xy'\,{\left( {xy + 1} \right)^2} + 3y\,{\left( {xy + 1} \right)^2} = 1 - 2yy'$ $collect\,\,like\,\,terms$ $3xy'\,{\left( {xy + 1} \right)^2} + 2yy' = 1 - 3y\,{\left( {xy + 1} \right)^2}$ $factor\,\,{y^,}$ $y' = \,\,\left[ {3x\,{{\left( {xy + 1} \right)}^2} + 2y} \right] = 1 - 3y\,{\left( {xy + 1} \right)^2}$ $solve\,\,for\,\,{y^,}$ $\frac{{dy}}{{dx}} = \frac{{1 - 3y\,{{\left( {xy + 1} \right)}^2}}}{{3x\,{{\left( {xy + 1} \right)}^2} + 2y}}$