## Calculus: Early Transcendentals (2nd Edition)

$y = \frac{1}{2}x$
$\begin{gathered} \cos \,\left( {x - y} \right) + \sin y = \sqrt 2 \,\,\,\,\,\,\,\,,\,\,\,\left( {\frac{\pi }{2},\frac{\pi }{4}} \right) \hfill \\ \hfill \\ use\,\,the\,\,implicit\,\,differentiation \hfill \\ \hfill \\ \frac{d}{{dx}}\,\,\left[ {\cos \,\left( {x - y} \right) + \sin y} \right] = \frac{d}{{dx}}\,\,\,\left[ {\sqrt 2 } \right] \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ - \sin \,\left( {x - y} \right)\,\left( {1 - {y^,}} \right) + \cos y \cdot {y^,} = 0 \hfill \\ \hfill \\ distribute \hfill \\ \hfill \\ {y^,}\sin \,\left( {x - y} \right) - \sin \,\left( {x - y} \right) + \cos y \cdot {y^,} = 0 \hfill \\ \hfill \\ factor\,\,{y^,} \hfill \\ \hfill \\ {y^,}\left( {\sin \,\left( {x - y} \right) + \cos y} \right) = \sin \,\left( {x - y} \right) \hfill \\ \hfill \\ solve\,\,for\,\,{y^,} \hfill \\ \hfill \\ {y^,} = \frac{{\sin \,\left( {x - y} \right)}}{{\sin \,\left( {x - y} \right)\cos y}} \hfill \\ \hfill \\ evaluate\,\,\,\,\left( {\frac{\pi }{2},\frac{\pi }{4}} \right)\,,\,so \hfill \\ \hfill \\ {y^,} = \frac{{\sin \,\left( {\frac{\pi }{2} - \frac{\pi }{4}} \right)}}{{\sin \,\left( {\frac{\pi }{2} - \frac{\pi }{4}} \right) + \cos \,\left( {\frac{\pi }{4}} \right)}} = \frac{1}{2} \hfill \\ \hfill \\ use\,\,the\,\,point\, - \,slope\,\,form \hfill \\ \hfill \\ y - {y_1} = m\,\left( {x - {x_1}} \right) \hfill \\ \hfill \\ then \hfill \\ \hfill \\ y - \frac{\pi }{4} = \frac{1}{2}\,\left( {x - \frac{\pi }{2}} \right) \hfill \\ \hfill \\ y - \frac{\pi }{2} = \frac{1}{2}x - \frac{\pi }{4} \hfill \\ \hfill \\ y = \frac{1}{2}x \hfill \\ \hfill \\ \end{gathered}$