Differentiation的簡單A.Maths.問題2

y = (π/ 2)sin(x + y)
dy/dx = {d[(π/ 2)sin(x + y)]/d(x+y)}{d(x + y)/dx}
dy/dx = (π/ 2)cos(x + y)(1 + dy/dx)
At (0, π/ 2),
dy/dx = (π/ 2)cos(0 + π/ 2 )(1 + dy/dx)
dy/dx = 0
This means the tangent is horizontal and the normal is vertical.
Horizontal tangent passing through (0, π/ 2) means the tangent is y = π/ 2.
Vertical normal passing through (0, π/ 2) means the normal is x = 0.