What is the equation of the line tangent to # f(x)=secxtanx # at # x=pi/3#?

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Answer 1 · 2015-12-30T21:22:22

#y-2sqrt3=20(x-pi/3)#

#f(pi/3)=2sqrt3#

The tangent line will pass through the point #(pi/3,2sqrt3)#.

The slope of the tangent line can be found through calculating #f'(pi/3)#.

To find #f'(x)#, use the product rule.

#f'(x)=tanxd/dx(secx)+secxd/dx(tanx)#

#=>tanx(secxtanx)+secx(sec^2x)#

#=>tan^2xsecx+sec^3x#

#f'(pi/3)=(sqrt3)^2(2)+(2)^3=20#

Relate the point #(pi/3,2sqrt3)# and slope of #20# in an equation in point-slope form.

#y-2sqrt3=20(x-pi/3)#