Calculus: Early Transcendentals (2nd Edition)

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