Hey everyone,

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.

Thanks!