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(1+x2x).
They are defined and continuous on interval (−∞,∞).
Since f′(x)=1+x21 and g′(x)=1−(1+x2x)21⋅1+x21+x2−1+x2x2=1+x21 then according to corollary arctan(x)=arcsin(1+x2x)+C on (−∞,∞).
To find constant plug any value of x, for example, x=0: arctan(0)=arcsin(1+020)+C or C=0.
So, we have the following fact: arctan(x)=arcsin(1+x2x) for all x.
Example 2. Consider functions f(x)=arctan(x) and g(x)=21arctan(1−x22x).
It can be easily proven that f′(x)=g′(x). However, function g(x) is not defined when x=±1. So, 21arctan(1−x22x)=arctan(x)+C on (−∞,−1),(−1,1),(1,∞).
It is interesting that constant will be different for different intervals.
For interval (−1,1) we plug x=0: 21arctan(1−022⋅0)=arctan(0)+C or C=0.
For interval (−∞,−1) we let x→−∞: 0=−2π+C or C=2π.
