# For 0<x<180 verify the monotony of the function xcosx+sin(-x).

### 2 Answers | Add Yours

We have the function f(x) = x*cos x + sin (-x)

f(x) = x*cos x - sin x

The first derivative of f(x) is f'(x)

=> f'(x) = cos x - x*(sin x) - cos x

=> f'(x) = -x*sin x

When 0< x< 180 we have sin x as positive, therefore the first derivative -x*sin x is negative, so the function is decreasing.

**Therefore the function is decreasing for the given range of x.**

The monotony of a function shows the behavior of the function: increasing or decreasing function.

A function is strictly increasing if it's first derivative is positive and it is decreasing if it's first derivative is negative.

We'll re-write the function, based on the fact that the sine function is odd:

f(x) = xcos x - sin x

We'll calculate the first derivative:

f'(x)= (xcos x – sin x)'

f'(x) = ( xcos x)'-( sin x)'

We notice that the first term is a product, so we'll apply the product rule:

f'(x) = x'*(cos x)+x*(cos x)' – cos x

f'(x)=1*cos x-x*sin x –cos x

We'll eliminate like terms:

f'(x)= -x*sin x

Since the sine function is positive over the range [0 ; 180], the values of x are positive and the product is negative, the first derivative is negative.

**The function y = f(x) = xcosx+sin(-x) is decreasing over the range [0, pi].**