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