What is the equation of the normal line of #f(x)= secxtanx# at #x = pi/8#?

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Answer 1 · 2017-04-16T01:15:19

#y-.488=1.454(x-pi/8)#

It helps to know some trig derivatives:

$tutorial.math.lamar.edu$

First, take the derivative of #f(x)#. You will need to use product rule :

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

Simplify:

#f'(x)=sec^3x+secxtan^2x#

Plug in #pi/8#

#f'(pi/8)=sec^3(pi/8)+sec(pi/8)tan^2(pi/8)~~1.454#

This is the slope of the tangent line but we need the slope of the normal line. Take the negative reciprocal of #1.454# to get #-.688#.

Now we need a point. Plug in #pi/8# into the initial equation:

#f(pi/8)=sec(pi/8)tan(pi/8)~~.448#

Now we have the point #(pi/8,.448)# and the slope #1.454#, plug into point-slope form:

#y-.488=1.454(x-pi/8)#