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.