Question

Please help me with this calculus 1 question

Answer 1

Answer:

tangent: y- $\frac{\pi }{4}$ = -4(x - $\frac{\pi }{2}$)

Step-by-step explanation:

differentiating

$y' sin(16x) + 16ycos(16x) = cos(2y) -2xy'sin(2y)$

isolating y' to one side and simplifying

$y'sin(16x) + 2xy'sin(2y) = cos(2y) - 16ycos(16x)$

$y'(sin(16x) + 2xsin(2y)) = cos(2y)-16ycos(16x)\\y' = \frac{cos(2y)-16ycos(16x)}{sin(16x) + 2xsin(2y)}$

when x = $\frac{\pi}{2}$,

y' = $\frac{cos(\frac{\pi }{2} ) - 4\pi cos(8\pi) }{sin(8\pi) + \pi sin(\frac{\pi }{2}) }$

y' = $\frac{0-4\pi }{0+\pi } = \frac{-4\pi }{\pi } =-4$

tangent: y- $\frac{\pi }{4}$ = -4(x - $\frac{\pi }{2}$)