I have this question,
“Find an equation for the line tangent to the graph y = sec x at (pi/3 , 2)”
So far I have that,
f(x) = sec x
f '(x) = sec x . tan x
I then subbed in x = (pi/3) into f '(x) to get the gradient which I found was 0, is this right?
I wanted to make I was approaching this question correctly and if I have that just means I sub m into the equation,
y - y1 = m(x - x1)
and finish it off.