find the tangent to y=27/(x^2+2) at (1,9)

Answer 1

y = 27(x^2+2)^-1
dy/dx = -27(2x)(x^2+2)^-2 = -54x/(x^2+2)^2
at x = 1 , dy/dx = -54/9 = -6

equation:
y = -6x+b , sub in the point (1,9)
9 = -6+b
b = 15

tangent equation: y = -6x + 15