How do you find the equation of the line tangent to #y=secx#, at (pi/3,2)?

1 Answer
Oct 24, 2017

#y=2sqrt3x-(2pisqrt3)/3+2#

Explanation:

Name the point #" "(pi/3,2)# by A.
#" "#
The line passes through the point #A(pi/3,2)" "#.
#" "#
The slope of this tangentbis determined by performing #y'# at
#" "#
#x=pi/3#
#" "#
Differentiating #y#
#" "#
#y' = (secx)'=tanxsecx#
#" "#
The slope of tangent line at #x=pi/3# is:
#" "#
#y'_(x=pi/3) =tan(pi/3)sec (pi/3)=sqrt3 xx 2=color (brown)(2sqrt3)#
#" "#
The equation of the tangent line with slope#" "color (brown)(2sqrt3)#
#" "#
passing through#" "A (pi/3,2)# is:
#" "#
#y-y_A= y_(x_A)(x-x_A)#
#" "#
#y-2=2sqrt3 (x-pi/3)#
#" "#
#y=2sqrt3x-(2pisqrt3)/3+2#