What is the slope of the line normal to the tangent line of #f(x) = secx+sin(2x-(3pi)/8) # at # x= (11pi)/8 #?

Question

What is the slope of the line normal to the tangent line of #f(x) = secx+sin(2x-(3pi)/8)  # at # x= (11pi)/8  #?

Emma Aldridge · Accepted Answer

The slope of the line normal to the tangent line
#m=1/((1+sqrt(2)/2)sqrt(2+sqrt2)+((3sqrt2)/2+1)sqrt(2-sqrt2)#
#m=0.18039870004873# From the given: #y=sec x+ sin (2x-(3pi)/8)# at #" "x=(11pi)/8# Take the first derivative #y'# #y'=sec x*tan x *(dx)/(dx)+ cos (2x-(3pi)/8)(2)(dx)/(dx)# Using #" "x=(11pi)/8# Take note: that by #color(Blue)("Half-Angle formulas")#, the following are obtained #sec ((11pi)/8)=-sqrt(2+sqrt2)-sqrt(2-sqrt2)# #tan ((11pi)/8)=sqrt2+1# and #2*cos (2x-(3pi)/8)=2*cos ((19pi)/8)# #=2*(sqrt2/4)(sqrt(2+sqrt2)-sqrt(2-sqrt2))# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ continuation #y'=(-sqrt(2+sqrt2)-sqrt(2-sqrt2))(sqrt2+1)# #+2*(sqrt2/4)(sqrt(2+sqrt2)-sqrt(2-sqrt2))# #y'=-(sqrt2+1)sqrt(2+sqrt2)-(sqrt2+1)sqrt(2-sqrt2)# #+(sqrt2)/2*sqrt(2+sqrt2)-sqrt2/2*sqrt(2-sqrt2)# further simplification #y'=(-1-sqrt2/2)sqrt(2+sqrt2)+((-3sqrt2)/2-1)sqrt(2-sqrt2)# For the normal line: # m=(-1)/(y')# #m=(-1)/((-1-sqrt2/2)sqrt(2+sqrt2)+((-3sqrt2)/2-1)sqrt(2-sqrt2))# #m=1/((1+sqrt2/2)sqrt(2+sqrt2)+((3sqrt2)/2+1)sqrt(2-sqrt2))# #m=0.180398700048733# God bless....I hope the explanation is useful.

Paisley Johnson · Answer

undefined The slope of the line normal to the tangent line of $ f(x) = \sec(x) + \sin\left(2x - \frac{3\pi}{8}\right) $ at $ x = \frac{11\pi}{8} $ is $ -\frac{1}{f'(x)} $, where $ f'(x) $ represents the derivative of $ f(x) $ with respect to $ x $. To find the slope, you would first find the derivative of the function $ f(x) $ and then evaluate it at $ x = \frac{11\pi}{8} $.