Find an equation of the tangent line to the curve at the given point.

Question

((pi/6),(2 square root of (3)/3))

y = sec (x)

Answer 1

y = secx
y' = secx tanx
at x = pi/6,
y' = 2/√3 * 1/√3 = 2/3

so, now we have a point and a slope, so the line is

y - 2/√3 = 2/3 (x-pi/6)

http://www.wolframalpha.com/input/?i=plot+y%3Dsec%28x%29+and+y+%3D+2%2F3+%28x-pi%2F6%29+%2B+++2%2F%E2%88%9A3