How do you find the equation for the tangent line to #sec^2x# at #x=pi/3#?

Question

William Abdalla · Accepted Answer

Equation of tangent is #8sqrt3x-y+4-(8pi)/3=0# The slope of tangent to a curve #y=f(x)# at a point #x=x_0# is given by #f'(x_0)#. Here the function is #y=sec^2x# and we are seeking tangent at point #x=pi/3# i.e. at #(pi/3,sec^2(pi/3))# or #(pi/3,4)#. As #f(x)=sec^2x#, #f'(x)=2secx xx secxtanx=2sec^2xtanx# and slope of tangent is #f'(pi/3)=2sec^2(pi/3)tan(pi/3)=8sqrt3# and hence equation of tangent is the equation of line passing through #(pi/3,4)# and slope #8sqrt3# i.e. #y-4=8sqrt3(x-pi/3)# or #8sqrt3x-y+4-(8pi)/3=0#

Axel Alvarez · Answer

undefined To find the equation for the tangent line to sec^2x at x=pi/3, we need to find the derivative of sec^2x and evaluate it at x=pi/3.

The derivative of sec^2x can be found using the chain rule.

The derivative of sec^2x is 2sec^2x * tanx.

Evaluating this derivative at x=pi/3, we have:

2sec^2(pi/3) * tan(pi/3).

Using the trigonometric identities, sec(pi/3) = 2 and tan(pi/3) = sqrt(3), we can simplify the equation to:

2(2)^2 * sqrt(3) = 8sqrt(3).

Therefore, the equation for the tangent line to sec^2x at x=pi/3 is y = 8sqrt(3)(x - pi/3) + sec^2(pi/3).