If $f:R \to R$ be defined by $f(x) = \frac{x}{{\sqrt {1 + {x^2}} }}$, then $({\rm{ }}fofof)(x) =$…………..

It is given that,, $f(x) = \frac{x}{{\sqrt {1 + {x^2}} }}$

$({\rm{ }}fofof)(x) =$ $= f[f\{ f(x)\} ]$

$= f\left[ {f\left( {\frac{x}{{\sqrt {1 + {x^2}} }}} \right)} \right] = f\left( {\frac{{\frac{x}{{\sqrt {1 + {x^2}} }}}}{{\sqrt {1 + \frac{{{x^2}}}{{1 + {x^2}}}} }}} \right)$

$= f\left[ {\frac{{x\sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^2}\left( {\sqrt {2{x^2}} + 1} \right)} }}} \right] = f\left( {\frac{x}{{\sqrt {1 + 2{x^2}} }}} \right)$

$= \frac{{\frac{x}{{\sqrt {1 + 2{x^2}} }}}}{{\sqrt {1 + \frac{{{x^2}}}{{1 + 2{x^2}}}} }} = \frac{{x\sqrt {1 + 2{x^2}} }}{{\sqrt {1 + 2{x^2}} \sqrt {1 + 3{x^2}} }}$

$= \frac{x}{{\sqrt {1 + 3{x^2}} }} = \frac{x}{{\sqrt {3{x^2} + 1} }}$