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

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:
$\frac{d}{\mathrm{dx}} \left[x \sec \left(x\right)\right] = \sec \left(x\right) + x \tan \left(x\right) \sec \left(x\right)$
Chain rule to derive second part:
$\frac{d}{\mathrm{dx}} \left[- \cos \left(2 x - \frac{\pi}{6}\right)\right] = 2 \sin \left(2 x - \frac{\pi}{6}\right)$
Put these together so the derivative is
$f ' \left(x\right) = \sec \left(x\right) + x \tan \left(x\right) \sec \left(x\right) + 2 \sin \left(2 x - \frac{\pi}{6}\right)$
Plug in $x = \frac{15 \pi}{8}$ 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 $0.2865$