Condition of Constancy of the Function

Fact. Suppose function `y=f(x)` is defined and continuous on interval `X` and has finite derivative `f'(x)`. Function `y=f(x)` is constant if and only if `f'(x)` for all `x` in `X`.

This fact means that if on some interval derivative of function equals 0 then function is constant their, its graph is just horizontal line.

Corollary. Suppose two function `y=f(x)` and `y=g(x)` are defined and continuous on interval `X`, and have finite derivatives `f'(x)` and `g(x)`. Function `y=f(x)` is constant if and only if `f'(x)` for all `x` in `X`. If `f'(x)=g'(x)` on interval `X` then `f(x)=g(x)+C`, where `C` is a constant for all `x` in `X`.

This corollary means that if functions have same derivatives on interval `X` then their difference is constant.

Example 1 . Consider functions `f(x)=arctan(x)` and `g(x)=arcsin(x/(sqrt(1+x^2)))`.

They are defined and continuous on interval `(-oo,oo)`.

Since `f'(x)=1/(1+x^2)` and `g'(x)=1/(1-(x/sqrt(1+x^2))^2)*(sqrt(1+x^2)-(x^2)/(sqrt(1+x^2)))/(1+x^2)=1/(1+x^2)` then according to corollary `arctan(x)=arcsin(x/(sqrt(1+x^2)))+C` on `(-oo,oo)`.

To find constant plug any value of `x`, for example, `x=0`: `arctan(0)=arcsin(0/sqrt(1+0^2))+C` or `C=0`.

So, we have the following fact: `arctan(x)=arcsin(x/sqrt(1+x^2))` for all `x`.

Example 2. Consider functions `f(x)=arctan(x)` and `g(x)=1/2 arctan((2x)/(1-x^2))`.

It can be easily proven that `f'(x)=g'(x)`. However, function `g(x)` is not defined when `x=+-1.` So, `1/2 arctan((2x)/(1-x^2))=arctan(x)+C` on `(-oo,-1),(-1,1),(1,oo)`.

It is interesting that constant will be different for different intervals.constancy of function

For interval `(-1,1)` we plug `x=0`: `1/2 arctan((2*0)/(1-0^2))=arctan(0)+C` or `C=0`.

For interval `(-oo,-1)` we let `x->-oo`: `0=-pi/2+C` or `C=pi/2`.

For interval `(1,oo)` we let `x->oo`: `0=pi/2+C` or `C=-pi/2`.

So,

`1/2arctan((2x)/(1-x^2))=arctan(x)+pi/2` on `(-oo,-1)`,

`1/2arctan((2x)/(1-x^2))=arctan(x)` on `(-1,1)`,

`1/2arctan((2x)/(1-x^2))=arctan(x)-pi/2` on `(1,oo)`.

Graph confirms these facts.