# What is the slope of the line normal to the tangent line of f(x) = sec^2x-xcos(x-pi/4)  at  x= (7pi)/4 ?

Jun 10, 2016

The reqd. slope = 4/((16+7pi).

#### Explanation:

$f \left(x\right) = {\sec}^{2} x - x \cos \left(x - \frac{\pi}{4}\right)$
$\therefore f ' \left(x\right) = 2 \cdot \sec x \cdot \sec x \cdot \tan x - \left\{x \cdot \left(- \sin \left(x - \frac{\pi}{4}\right)\right) + \cos \left(x - \frac{\pi}{4}\right)\right\} = 2 {\sec}^{2} x \cdot \tan x + x \cdot \sin \left(x - \frac{\pi}{4}\right) - \cos \left(x - \frac{\pi}{4}\right)$

But $f ' \left(x\right)$ is the slope tangent (tgt.) line at $\left(x , f \left(x\right)\right)$
Hence, slope of the tangent line at $x = \frac{7 \pi}{4}$ is
$f ' \left(\frac{7 \pi}{4}\right) = 2 {\sec}^{2} \left(\frac{7 \pi}{4}\right) \tan \left(\frac{7 \pi}{4}\right) + \left(\frac{7 \pi}{4}\right) \sin \left(\frac{7 \pi}{4} - \frac{\pi}{4}\right) - \cos \left(\frac{7 \pi}{4} - \frac{\pi}{4}\right) = 2 \cdot 2 \cdot \left(- 1\right) + \left(\frac{7 \pi}{4}\right) \cdot \left(- 1\right) - 0 = - 4 - \frac{7 \pi}{4} = - \frac{16 + 7 \pi}{4}$
As normal is perpendicular to tgt. line, the reqd. slope = 4/((16+7pi).