Question

No 1 dan 2 caranya kak...

Answer 1

1. x².y + x².y³ + x.y - 3x² + 6y + 10 = 0
    x².(y+y³-3) + y.(x+6) + 10 = 0
    2x.(y+y³-3) dx + x².(1+3y²) dy + (x+6) dy + y dx = 0

    (x².(1+3y²) + (x+6)) dy = -(2x.(y+y³-3) + y) dx
                              dy/dx = -(2x.(y+y³-3) + y)/(x².(1+3y²) + (x+6))

2. lim(x→0) (tan (x/2)/(2x.cos x)
    = lim(x→0) (tan (x/2)/(2x).(1/cos x)
    = lim(x→0) (1/4).(1/cos x)
    = (1/4).(1/cos 0)
    = (1/4).(1)
    = 1/4