# How do you find the intervals of increasing and decreasing using the first derivative given y=sinx+cosx?

Jul 25, 2017

The function is increasing on intervals of the form $\left[- \frac{3 \pi}{4} + 2 n \pi , \frac{\pi}{4} + 2 n \pi\right]$ and decreasing on intervals of the form $\left[\frac{\pi}{4} + 2 n \pi , \frac{5 \pi}{4} + 2 n \pi\right]$, where $n = 0 , \setminus \pm 1 , \setminus \pm 2 , \ldots$

#### Explanation:

The derivative of $y = \sin x + \cos x$ is $\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x - \sin x$. Setting this equal to zero yields $\cos x = \sin x$ (which is equivalent to $\tan x = 1$). This occurs when $x = \ldots , - \frac{7 \pi}{4} , - \frac{3 \pi}{4} , \frac{\pi}{4} , \frac{5 \pi}{4} , \ldots$.

You can check that $\cos x > \sin x$ (so $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$) which occurs when $x \setminus \in \left(- \frac{3 \pi}{4} , \frac{\pi}{4}\right)$, when $x \setminus \in \left(\frac{5 \pi}{4} , \frac{9 \pi}{4}\right)$, etc...).

You can check that $\cos x < \sin x$ (so $\frac{\mathrm{dy}}{\mathrm{dx}} < 0$) which occurs when $x \setminus \in \left(\frac{- 7 \pi}{4} , - \frac{3 \pi}{4}\right)$, when $x \setminus \in \left(\frac{\pi}{4} , \frac{5 \pi}{4}\right)$, etc...).

The answer above follows from these observations.

You can confirm this visually by looking at the graph.

graph{sin(x)+cos(x) [-10, 10, -5, 5]}