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

Question

y=tan x, (pi/4,1)

Any help would be great. Thanks!

Answer 1

slope = y' = sec^2(x)
y'(pi/4) = 2

y - y1 = y'(x - x1)

y - 1 = 2(x - pi/4)

y - 1 = 2x - pi/2

y - 2x = 1 - pi/2

Answer 2

first you must differentiate this function
then sub in the x coordinate to get the slope of the tangent at that point then use Y-y1=m(X-x1) and you have your equation

Answer 3

dy/dx = sec² x = 1 / cos ² x
1 / cos ² (π/4) = 1 /(1 / √2)² = 2 = m
y - b = m (x - a)
y - 1 = 2 (x - π/4)
y = 2 x - π/2 + 1