# How do you find the intervals of increasing and decreasing using the first derivative given y=cos^2x-sin^2x?

##### 1 Answer
Dec 1, 2016

$y = {\cos}^{2} x - {\sin}^{2} x$

$y = \left(\cos x + \sin x\right) \left(\cos x - \sin x\right)$

By the product rule:

$y ' = \left(\cos x - \sin x\right) \left(\cos x - \sin x\right) + \left(\cos x + \sin x\right) \left(- \sin x - \cos x\right)$

$y ' = {\cos}^{2} x - 2 \sin x \cos x + {\sin}^{2} x + \left(- \cos x \sin x - {\sin}^{2} x - {\cos}^{2} x - \cos x \sin x\right)$

$y ' = {\cos}^{2} x + {\sin}^{2} x - 2 \sin x \cos x + \left(- \left(\cos x \sin x + {\sin}^{2} x + {\cos}^{2} x + \cos x \sin x\right)\right)$

$y ' = 1 - 2 \sin x \cos x + \left(- \left(2 \cos x \sin x + 1\right)\right)$

$y ' = 1 - 2 \sin x \cos x - 2 \sin x \cos x - 1$

$y ' = - 4 \sin x \cos x$

$y ' = - 2 \left(2 \sin x \cos x\right)$

$y ' = - 2 \sin 2 x$

The function will be increasing when $y ' > 0$ and decreasing when $y < 0$.

We set the derivative to $0$ and solve for $x$ first and then use test points to determine the intervals.

$0 = - 2 \sin 2 x$

$0 = - 4 \sin x \cos x$

$x = 0 , \frac{\pi}{2} , \pi , \frac{3 \pi}{2} , 2 \pi$

We try $\frac{\pi}{4} , \frac{3 \pi}{4} , \frac{5 \pi}{4} , \frac{7 \pi}{4}$ as test points. The results are as follows, with the function being $f \left(x\right)$ and the derivative $f ' \left(x\right)$

$f ' \left(\frac{\pi}{4}\right) = - 4 \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right) = -$

$f ' \left(\frac{3 \pi}{4}\right) = - 4 \sin \left(\frac{3 \pi}{4}\right) \cos \left(\frac{3 \pi}{4}\right) = +$

$f ' \left(\frac{5 \pi}{4}\right) = - 4 \sin \left(\frac{5 \pi}{4}\right) \cos \left(\frac{5 \pi}{4}\right) = -$

f'((7pi)/4)) = -4sin((7pi)/4)cos((7pi)/4) = +

Hence, in the interval 0 ≤ x ≤ 2pi, the intervals of increase/decrease are:

•Decreasing over 0 ≤ x ≤ pi/2 and pi ≤ x ≤ (3pi)/2.

•Increasing over pi/2 ≤ x ≤ pi and (3pi)/2 ≤ x ≤ 2pi

Hopefully this helps!