# What is the equation of the normal line of f(x)= secxtanx at x = pi/8?

Apr 16, 2017

$y - .488 = 1.454 \left(x - \frac{\pi}{8}\right)$

#### Explanation:

It helps to know some trig derivatives:

First, take the derivative of $f \left(x\right)$. You will need to use product rule :

$f ' \left(x\right) = \sec x \left({\sec}^{2} \left(x\right)\right) + \tan x \left(\sec x \tan x\right)$

Simplify:

$f ' \left(x\right) = {\sec}^{3} x + \sec x {\tan}^{2} x$

Plug in $\frac{\pi}{8}$

$f ' \left(\frac{\pi}{8}\right) = {\sec}^{3} \left(\frac{\pi}{8}\right) + \sec \left(\frac{\pi}{8}\right) {\tan}^{2} \left(\frac{\pi}{8}\right) \approx 1.454$

This is the slope of the tangent line but we need the slope of the normal line. Take the negative reciprocal of $1.454$ to get $- .688$.

Now we need a point. Plug in $\frac{\pi}{8}$ into the initial equation:

$f \left(\frac{\pi}{8}\right) = \sec \left(\frac{\pi}{8}\right) \tan \left(\frac{\pi}{8}\right) \approx .448$

Now we have the point $\left(\frac{\pi}{8} , .448\right)$ and the slope $1.454$, plug into point-slope form:

$y - .488 = 1.454 \left(x - \frac{\pi}{8}\right)$