differentiation

Let (x_1, 6) be the point of the tangent to the curve.
Slope of the straight line = 0
dy/dx|(x_1,6) = 0
-4(x_1)/((x_1)^2 + 1)^2 = 0
-4x_1 = 0
x_1 = 0
The coordinate of the point of tangent to the curve = (0, 6)
dy/dx = -4x/(x^2 + 1)^2
y = ∫-4x/(x^2 + 1)^2 dx
= -2 ∫ 1/(x^2 + 1)^2 d(x^2 + 1)
= 2 [1/(x^2 + 1)] + C
= 2/(x^2 + 1) + C
Put x = 0 and y = 6,
6 = 2/(0 + 1) + C
C = 4
The required equation:
y = 2/(x^2 + 1) + 4