Answer
\[y = \frac{5}{{1 + 2\pi }}x - \frac{{{\pi ^2}}}{{1 + 2\pi }} + \pi \]
Work Step by Step
\[\begin{gathered}
\sin y + 5x = {y^2}\,\,\,\,\,\,\,\,,\,\,\,\,\,\left( {\frac{{{\pi ^2}}}{5},\pi } \right) \hfill \\
\hfill \\
use\,\,the\,\,implicit\,\,differentiation \hfill \\
\hfill \\
\cos y{y^,} + 5 = 2y{y^,} \hfill \\
\hfill \\
collect\,\,like\,\,terms \hfill \\
\hfill \\
{y^,}\cos y - 2y{y^2} = - 5 \hfill \\
\hfill \\
factor\,\,{y^,} \hfill \\
\hfill \\
{y^,}\left( {\cos y - 2y} \right) = - 5 \hfill \\
\hfill \\
solve\,for\,\,{y^,} \hfill \\
\hfill \\
{y^,} = - \frac{5}{{\cos y - 2y}} \hfill \\
\hfill \\
evaluate\,\,\,\,\left( {\frac{{{\pi ^2}}}{5},\pi } \right) \hfill \\
\hfill \\
{y^,} = - \frac{5}{{\cos \pi - 2\pi }} = - \frac{5}{{ - 1 - 2\pi }} \hfill \\
\hfill \\
so \hfill \\
use\,\,the\,\,point\, - \,slope\,\,form \hfill \\
\hfill \\
y - \pi = \frac{5}{{1 + 2\pi }}\,\left( {x - \frac{{{\pi ^2}}}{5}} \right) \hfill \\
\hfill \\
y - \pi = \frac{5}{{1 + 2\pi }}x - \frac{{{\pi ^2}}}{{1 + 2\pi }} \hfill \\
\hfill \\
y = \frac{5}{{1 + 2\pi }}x - \frac{{{\pi ^2}}}{{1 + 2\pi }} + \pi \hfill \\
\hfill \\
\end{gathered} \]
