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

Jun 9, 2018

The slope of the normal line is given by $\frac{1}{3 \sqrt{2 + \sqrt{2}}}$

#### Explanation:

The derivative is given by

$f ' \left(x\right) = 2 \sin \left(\frac{\pi}{8} + 2 x\right) + \sec \left(x\right) \cdot \tan \left(x\right)$

for $x = \frac{11}{8} \pi$ we get

$f ' \left(\frac{\pi}{8}\right) = \sqrt{2 - \sqrt{2}} - 2 \frac{\sqrt{2 + \sqrt{2}}}{2 - \sqrt{2}}$
simplified to

$f ' \left(\frac{\pi}{8}\right) = - 3 \sqrt{2 + \sqrt{2}}$

so the slope of the normal line is given by

$f ' \left(\frac{\pi}{8}\right) = \frac{1}{3 \sqrt{2 + \sqrt{2}}}$