# Condition of Constancy of the Function

Fact. Suppose function ${y}={f{{\left({x}\right)}}}$ is defined and continuous on interval ${X}$ and has finite derivative ${f{'}}{\left({x}\right)}$. Function ${y}={f{{\left({x}\right)}}}$ is constant if and only if ${f{'}}{\left({x}\right)}$ 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 ${y}={f{{\left({x}\right)}}}$ and ${y}={g{{\left({x}\right)}}}$ are defined and continuous on interval ${X}$, and have finite derivatives ${f{'}}{\left({x}\right)}$ and ${g{{\left({x}\right)}}}$. Function ${y}={f{{\left({x}\right)}}}$ is constant if and only if ${f{'}}{\left({x}\right)}$ for all ${x}$ in ${X}$. If ${f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)}$ on interval ${X}$ then ${f{{\left({x}\right)}}}={g{{\left({x}\right)}}}+{C}$, 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 ${f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}$.

They are defined and continuous on interval ${\left(-\infty,\infty\right)}$.

Since ${f{'}}{\left({x}\right)}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$ and ${g{'}}{\left({x}\right)}=\frac{{1}}{{{1}-{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}^{{2}}}}\cdot\frac{{\sqrt{{{1}+{{x}}^{{2}}}}-\frac{{{{x}}^{{2}}}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}}}{{{1}+{{x}}^{{2}}}}=\frac{{1}}{{{1}+{{x}}^{{2}}}}$ then according to corollary ${\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{{\sqrt{{{1}+{{x}}^{{2}}}}}}\right)}}}+{C}$ on ${\left(-\infty,\infty\right)}$.

To find constant plug any value of ${x}$, for example, ${x}={0}$: ${\operatorname{arctan}{{\left({0}\right)}}}={\operatorname{arcsin}{{\left(\frac{{0}}{\sqrt{{{1}+{{0}}^{{2}}}}}\right)}}}+{C}$ or ${C}={0}$.

So, we have the following fact: ${\operatorname{arctan}{{\left({x}\right)}}}={\operatorname{arcsin}{{\left(\frac{{x}}{\sqrt{{{1}+{{x}}^{{2}}}}}\right)}}}$ for all ${x}$.

Example 2. Consider functions ${f{{\left({x}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ and ${g{{\left({x}\right)}}}=\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}$.

It can be easily proven that ${f{'}}{\left({x}\right)}={g{'}}{\left({x}\right)}$. However, function ${g{{\left({x}\right)}}}$ is not defined when ${x}=\pm{1}.$ So, $\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+{C}$ on ${\left(-\infty,-{1}\right)},{\left(-{1},{1}\right)},{\left({1},\infty\right)}$.

It is interesting that constant will be different for different intervals.

For interval ${\left(-{1},{1}\right)}$ we plug ${x}={0}$: $\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}\cdot{0}}}{{{1}-{{0}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({0}\right)}}}+{C}$ or ${C}={0}$.

For interval ${\left(-\infty,-{1}\right)}$ we let ${x}\to-\infty$: ${0}=-\frac{\pi}{{2}}+{C}$ or ${C}=\frac{\pi}{{2}}$.

For interval ${\left({1},\infty\right)}$ we let ${x}\to\infty$: ${0}=\frac{\pi}{{2}}+{C}$ or ${C}=-\frac{\pi}{{2}}$.

So,

$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}+\frac{\pi}{{2}}$ on ${\left(-\infty,-{1}\right)}$,

$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}$ on ${\left(-{1},{1}\right)}$,

$\frac{{1}}{{2}}{\operatorname{arctan}{{\left(\frac{{{2}{x}}}{{{1}-{{x}}^{{2}}}}\right)}}}={\operatorname{arctan}{{\left({x}\right)}}}-\frac{\pi}{{2}}$ on ${\left({1},\infty\right)}$.

Graph confirms these facts.