question on finding the slope of a line!?

Question

how do u find the slope of the line normal to the graph of y=3secx at x=pi/4?

Answer 1

f(x) = 3 sec x
f `(x) = 3 sec x tan x
f `( π/4 ) = 3 sec (π/4) tan (π/4)
f `(π/4) = 3 (1 / cos π/4) tan(π/4)
f `(π/4) = 3 √2

m = Slope of normal = - 1 / (3 √2)
m = - √ 2 / 6

Answer 2

y' = 3secxtanx = 3*1/sqrt(2)* 1 = 3sqrt(2)/2
So slope of normal is -2/(3sqrt(2)) = -sqrt(2)/3