Answer
\[y' = \frac{{1 - y\cos xy}}{{x\cos xy - 1}}\]
Work Step by Step
\[find\,\,\frac{{dy}}{{dx}}\,\,\,\,using\,\,implicit\,\,\,differentitation.\]
\[\cos xy\, \cdot {\left( {xy} \right)^\prime } = 1 + y'\]
\[Using\,\,the\,\,product\,\,rule\]
\[\cos xy \cdot \,\left( {y + xy'} \right) = 1 + y'\]
\[multiply\]
\[\cos xy \cdot xy' - y' = 1 - \cos xy \cdot y\]
\[factor\]
\[y'\,\left( {x\cos xy - 1} \right) = 1 - y\cos xy\]
\[solve\,\,for\,\,{y^{\,,}}\]
\[y' = \frac{{1 - y\cos xy}}{{x\cos xy - 1}}\]
