# What is the slope of the line normal to the tangent line of f(x) = sec^2x-xcos(x-pi/4)  at  x= (15pi)/8 ?

Feb 16, 2018

$\implies y = 0.063 \left(x - \frac{15 \pi}{8}\right) - 1.08$

Interactive graph

#### Explanation:

The first thing we'll need to do is calculate $f ' \left(x\right)$ at $x = \frac{15 \pi}{8}$.

Let's do this term by term. For the ${\sec}^{2} \left(x\right)$ term, note that we have two functions embedded within one another: ${x}^{2}$, and $\sec \left(x\right)$. So, we'll need to use a chain rule here:

$\frac{d}{\mathrm{dx}} {\left(\sec \left(x\right)\right)}^{2} = 2 \sec \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(\sec \left(x\right)\right)$

$\textcolor{b l u e}{= 2 {\sec}^{2} \left(x\right) \tan \left(x\right)}$

For the 2nd term, we'll need to use a product rule. So:

$\frac{d}{\mathrm{dx}} \left(x \cos \left(x - \frac{\pi}{4}\right)\right) = \textcolor{red}{\frac{d}{\mathrm{dx}} \left(x\right)} \cos \left(x - \frac{\pi}{4}\right) + \textcolor{red}{\frac{d}{\mathrm{dx}} \cos \left(x - \frac{\pi}{4}\right)} \left(x\right)$

$\textcolor{b l u e}{= \cos \left(x - \frac{\pi}{4}\right) - x \sin \left(x - \frac{\pi}{4}\right)}$

You may wonder why we didn't use a chain rule for this part, since we have an $\left(x - \frac{\pi}{4}\right)$ inside the cosine. The answer is we implicitly did, but we ignored it. Notice how the derivative of $\left(x - \frac{\pi}{4}\right)$ is simply 1? Hence, multiplying that on doesn't change anything, so we do not write it out in calculations.

Now, we put everything together:

$\frac{d}{\mathrm{dx}} \left({\sec}^{2} x - x \cos \left(x - \frac{\pi}{4}\right)\right) = \textcolor{v i o \le t}{2 {\sec}^{2} \left(x\right) \tan \left(x\right) - \cos \left(x - \frac{\pi}{4}\right) + x \sin \left(x - \frac{\pi}{4}\right)}$

Now, we need to find the slope of the line tangent to $f \left(x\right)$ at $x = \frac{15 \pi}{8}$. To do this, we just plug this value into $f ' \left(x\right)$:

$f ' \left(\frac{15 \pi}{8}\right) = \left(2 {\sec}^{2} \left(\frac{15 \pi}{8}\right) \tan \left(\frac{15 \pi}{8}\right) - \cos \left(\frac{15 \pi}{8} - \frac{\pi}{4}\right) + \frac{15 \pi}{8} \sin \left(\frac{15 \pi}{8} - \frac{\pi}{4}\right)\right) = \textcolor{v i o \le t}{\approx - 6.79}$

However, what we want is not the line tangent to f(x), but the line normal to it. To get this, we just take the negative reciprocal of the slope above.

m_(norm) = -1/-15.78 color(violet)(~~0.015)

Now, we just fit everything into point slope form:

#y = m(x-x_0) + y_0

$\implies y = 0.063 \left(x - \frac{15 \pi}{8}\right) - 1.08$

Take a look at this interactive graph to see what this looks like!

Hope that helped :)