Answer:
tangent: y- [tex]\frac{\pi }{4}[/tex] = -4(x - [tex]\frac{\pi }{2}[/tex])
Step-by-step explanation:
differentiating
[tex]y' sin(16x) + 16ycos(16x) = cos(2y) -2xy'sin(2y)[/tex]
isolating y' to one side and simplifying
[tex]y'sin(16x) + 2xy'sin(2y) = cos(2y) - 16ycos(16x)[/tex]
[tex]y'(sin(16x) + 2xsin(2y)) = cos(2y)-16ycos(16x)\\y' = \frac{cos(2y)-16ycos(16x)}{sin(16x) + 2xsin(2y)}[/tex]
when x = [tex]\frac{\pi}{2}[/tex],
y' = [tex]\frac{cos(\frac{\pi }{2} ) - 4\pi cos(8\pi) }{sin(8\pi) + \pi sin(\frac{\pi }{2}) }[/tex]
y' = [tex]\frac{0-4\pi }{0+\pi } = \frac{-4\pi }{\pi } =-4[/tex]
tangent: y- [tex]\frac{\pi }{4}[/tex] = -4(x - [tex]\frac{\pi }{2}[/tex])
