# Question 5c275

May 15, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 \pi + \left(\frac{1}{2}\right)$

#### Explanation:

Use product rule :

$\frac{\mathrm{dy}}{\mathrm{dx}} = x \left({\cos}^{2} \left(x\right)\right) ' + \left({\cos}^{2} \left(x\right)\right) \left(x\right) '$

We will need to use the chain rule to find the derivative of cos^2(x))#:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - x \left(2 \cos \left(x\right) \sin \left(x\right)\right) + {\cos}^{2} x$

Plug in $\frac{\pi}{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(\frac{\pi}{4}\right) \left(2 \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{4}\right)\right) + {\cos}^{2} \left(\frac{\pi}{4}\right)$

This becomes:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \left(\frac{\pi}{4}\right) \left(2 \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{2}}{2}\right)\right) + \left(\frac{2}{4}\right)$

Evaluate:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\pi}{4} + \left(\frac{1}{2}\right)$