Question

Please help me with this calculus 1 question

Answer 1

tangent: y-
`(\pi )/(4)` = -4(x -
`(\pi )/(2)`)

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' = (cos(2y)-16ycos(16x))/(sin(16x) + 2xsin(2y))`

when x =
`(\pi)/(2)`,

y' =
`(cos((\pi )/(2) ) - 4\pi cos(8\pi) )/(sin(8\pi) + \pi sin((\pi )/(2)) )`

y' =
`(0-4\pi )/(0+\pi ) = (-4\pi )/(\pi ) =-4`

tangent: y-
`(\pi )/(4)` = -4(x -
`(\pi )/(2)`)