# How do you find the intervals of increasing and decreasing using the first derivative given y=sin^2x+sinx in 0 ≤ x ≤ (5pi)/2?

Jan 7, 2017

Differentiate:

$y = \sin x \left(\sin x + 1\right)$

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

$y ' = \cos x \sin x + \cos x + \sin x \cos x$

$y ' = 2 \sin x \cos x + \cos x$

$y ' = \sin 2 x + \cos x$

The intervals of increase/decrease will be obtained by finding the signs of the derivative. To do this though, we will have to find the critical numbers of the function. This derivative is defined for all $x$ values.

$0 = \sin 2 x + \cos x$

$0 = 2 \sin x \cos x + \cos x$

$0 = \cos x \left(2 \sin x + 1\right)$

$x = \frac{\pi}{2} , \frac{3 \pi}{2} , \frac{7 \pi}{6} , \frac{11 \pi}{6}$, in 0 ≤ x ≤ 2pi.

So, at each of these points, the function will change directions.

Select test points between each of these numbers. If $f ' \left(x\right) > 0$, then $f \left(x\right)$ is increasing. If $f \left(x\right) < 0$, then $f ' \left(x\right)$ is decreasing.

Test point 1: $\pi$

$y ' = \sin \left(2 \pi\right) + \cos \left(\pi\right)$

$y ' = 0 + - 1$

$y ' = - 1 < 0$

$\therefore$ $f \left(x\right)$ is decreasing in $\frac{\pi}{2} < x < \frac{7 \pi}{6}$

Test point 2: $\frac{5 \pi}{4}$

$y ' = \sin \left(\frac{5 \pi}{2}\right) + \cos \left(\frac{5 \pi}{4}\right)$

$y ' = 1 - \frac{1}{\sqrt{2}}$

$y ' = \frac{\sqrt{2} - 1}{\sqrt{2}} > 0$

$\therefore$$f \left(x\right)$ is increasing in $\frac{7 \pi}{6} < x < \frac{3 \pi}{2}$

Test point 3: $\frac{7 \pi}{4}$

$y ' = \sin \left(2 \frac{7 \pi}{4}\right) + \cos \left(\frac{7 \pi}{4}\right)$

$y ' = \sin \left(\frac{7 \pi}{2}\right) + \cos \left(\frac{7 \pi}{4}\right)$

$y ' = - 1 + \frac{1}{\sqrt{2}}$

$y ' = \frac{1 - \sqrt{2}}{\sqrt{2}} < 0$

$\therefore$$f \left(x\right)$ is decreasing in $\frac{3 \pi}{2} < x < \frac{11 \pi}{6}$

Since trig functions are periodic, we have to link back from $\frac{11 \pi}{6}$ and $\frac{5 \pi}{2}$.

Test point 4: $\left(\frac{\pi}{6}\right)$

$y ' = \sin \left(2 \left(\frac{\pi}{6}\right)\right) + \cos \left(\frac{\pi}{6}\right)$

$y ' = \sin \left(\frac{\pi}{3}\right) + \cos \left(\frac{\pi}{6}\right)$

$y ' = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}$

$y ' = \frac{2 \sqrt{3}}{2}$

$y ' = \sqrt{3} > 0$

$\therefore$$f \left(x\right)$ is increasing in the interval (11pi)/6 ≤ x ≤ (5pi)/2

Hopefully this helps!