Question

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

Answer 1

The function is increasing on intervals of the form `[-(3pi)/4+2npi,pi/4+2npi]` and decreasing on intervals of the form `[pi/4+2npi,(5pi)/4+2npi]`, where `n=0,\pm 1,\pm 2,...`

Explanation:

The derivative of `y=sin x+cos x` is `dy/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=...,-(7pi)/4,-(3pi)/4,pi/4,(5pi)/4,...`.

You can check that `cos x>sin x` (so `dy/dx>0`) which occurs when `x\in (-(3pi)/4,pi/4)`, when `x\in ((5pi)/4,(9pi)/4)`, etc...).

You can check that `cos x < sin x` (so `dy/dx<0`) which occurs when `x\in ((-7pi)/4,-(3pi)/4)`, when `x\in (pi/4,(5pi)/4)`, 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]}