What is the slope of the line normal to the tangent line of #f(x) = xsecx-cos(2x-pi/6) # at # x= (15pi)/8 #?

1 Answer
Bilal N.
Feb 19, 2018

0.2865...

Explanation:

So first derive the equation, then plug in x into the derivative, then find the negative reciprocal of that slope to get the normal slope.

Use product rule to derive the first part:
#d/dx[xsec(x)]= sec(x)+ xtan(x)sec(x)#
Chain rule to derive second part:
#d/dx[-cos(2x-pi/6)]=2sin(2x-pi/6)#
Put these together so the derivative is
#f'(x)= sec(x)+ xtan(x)sec(x)+2sin(2x-pi/6)#
Plug in #x=(15pi)/8# to get the slope of the parallel line as #-3.49041#.
The slope of the normal (perpendicular) is the negative reciprocal of that slope, which will be #0.2865#

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